Tight Hardness Results for Some Approximation Problems [mostly Håstad]

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Tight Hardness Results for Some Approximation Problems [mostly Håstad]. Adi Akavia Dana Moshkovitz S. Safra. “Road-Map” for Chapter I. Gap-3-SAT . expander. Gap-3-SAT-7 . Parallel repetition lemma. par[  , k] .  X  Y is NP-hard. Maximum Satisfaction. Def : Max-SAT Instance : - PowerPoint PPT Presentation

Transcript of Tight Hardness Results for Some Approximation Problems [mostly Håstad]

  • Tight Hardness Results forSome Approximation Problems

    [mostly Hstad]

    Adi Akavia Dana Moshkovitz S. Safra

  • Road-Map for Chapter IParallel repetition lemmaXY is NP-hardexpanderpar[, k] Gap-3-SAT-7 Gap-3-SAT

  • Maximum SatisfactionDef: Max-SATInstance:A set of variables Y = { Y1, , Ym }A set of Boolean-functions (local-tests) over Y = { 1, , l } Maximization:We define () = maximum, over all assignments to Y, of the fraction of i satisfied Structure:Various versions of SAT would impose structure properties on Y, Ys range and

  • Max-E3-Lin-2Def: Max-E3-Lin-2Instance: 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)

  • ExampleAssigning x1-6=1, x7-9=0 satisfy all but the third equation.No assignment can satisfy all equation, as the sum of all leftwing of equations equals zero (every variable appears twice) while the rightwing sums to 1.Therefore, (L)=5/6.x1+x2+x3=1 (mod 2)x4+x5+x6=1x7+x8+x9=1x1+x4+x7=0x2+x5+x8=0x3+x6+x9=0

  • 2-Variables Functional SATDef[ XY ]: overvariables X,Y of range Rx,Ry respectivelyeach is of the form xy: RxRyan assignment A(xRx,yRy) satisfies xy iff xy(A(x))=A(y)[ Namely, every value to X determines exactly 1 satisfying value for Y] Thm: distinguishing between A satisfies all A satisfies -0(1)

  • Proof OutlineDef: 3SAT is SAT where every i is a disjunction of 3 literals.

    Def: gap-3SAT-7 is gap-3SAT with the additional restriction, that every variable appears in exactly 7 local-tests

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

    XY is NP-hardpar[, k]Gap-3-SAT-7Gap-3-SAT?

  • ExpandersDef: a graph G(V,E) is a c-expander if for every SV, |S| |V|: |N(S)\S| c|S|[where N(S) denotes the set of neighbors of S] Lemma: For every m, one can construct in poly-time a 3-regular, m-vertices, c-expander, for some constant c>0 Corollary: a cut between S and V\S, for |S| |V| must contain > c|S| edgesXY is NP-hardpar[, k]Gap-3-SAT-7Gap-3-SATExpanders?

  • Reduction Using ExpandersAssume for which () is either 1 or 1-20/c. is with the following changes:an occurrence of y in i is replaced by a variable xy,i Let Gy, for every y, be a 3-regular, c-expander over all occurrences xy,i of yFor every edge connecting xy,i to xy,j in Gy, add to the clauses (xy,i xy,j) and (xy,i xy,j) It is easy to see that: || 10 ||Each variable xy,i of appears in exactly 7 i ensuring equalityconstructible by the Lemma

  • Correctness of the Reduction is completely satisfiable iff isIn case is unsatisfiable: () < 1-20/c Let A be an optimal assignment to Let Amaj assign xy,i the value assigned by A to the majority, over j, of variables xy,j Let FA and FAmaj be the sets of unsatisfied by A and Amaj respectively: ||(1-()) = |FA| = |FAFAmaj|+|FA\FAmaj| |FAFAmaj|+c|FAmaj\FA| c|FAmaj| and since Amaj is in fact an assignment to () 1- c(1- ())/10 < 1- c(20/c)/10= 1-

  • NotationsDef: For a 3SAT formula over Boolean variables Z, Let Zk be the set of all k-sequences of s variablesLet k be the set of all k-sequences of s clauses

    Def: For any VYk and Ck, let RY be the set of all assignments to V RX be the set of all satisfying assignments to C

    Def: For any set of k variables VZk, and a set of k clauses Ck, denote V C V is a choice of one variable of each clause in C.

  • Parallel SATDef: for a 3SAT formula over Boolean variables Z, define par[, k]:par[, k] has two types of variables:yV for every set VYk, where yVs range is the set RY of all assignments to VxC for every set Ck, where xCs range is the set RX of all satisfying assignments to all clauses in Cpar[, k] has a local-test [C,V] for each V C which accepts if xCs value restricted to V is yVs value (namely, if the assignments to T[C] and T[V] are consistent)|RY|=2k|RX|=7k

  • Gap Increases with kNote that if () = 1 then (par[, k]) = 1 On the other hand, if is not satisfiable:Lemma: (par[, k]) ()ck for some c>0 Proof: first note that 1-(par[, 1]) (1-())/3now, to prove the lemma, apply the Parallel-Repetition lemma [Raz] to par[, 1]In any assignment to s variables, any unsatisfied clause in induces at least 1 (out of corresponding 3) unsatisfied par[, 1] XY is NP-hardpar[, k]Gap-3-SAT-7Gap-3-SATParallel repetition lemma

  • Conclusion: XY is NP-hardDenote: = par[, k]X={xC}Y={yV}Then,

    distinguish between: A satisfies all A satisfies -0(1)

  • Road-Map for Chapter II() = () LLC-Lemma: (L) = +/2 (par[,k]) > 42Long codeL XY is NP-hard

  • Main TheoremThm: gap-Max-E3-Lin-2(1-, +) is NP-hard. That is, for every constant 0
  • 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).

  • Distributional AssignmentsLet be a SAT instance over variables Z of range R.Let (R) be all distributions over R Def: a distributional-assignment to is A: Z (R) Denote by () the maximum over distributional-assignments A of the average probability for to be satisfied, if variables` values are chosen according to A Clearly () (). Moreover Prop: () ()

  • Distributional-assignment to 11110000x1x2x3xn11110000x1x2x3xnOR:

  • Restriction and ExtensionDef: For any yY over RY and xX over RX s.t xyThe natural restriction of an aRX to Ry is denoted a|yThe elevation of a subset FP[RY] to RX is the subset F*P[RX] of all members a of RX for which xy(a) F F* = { a | a|y F }

  • Long-CodeIn the long-code the set of legal-words consists of all monotone dictatorshipsThis is the most extensive binary code,as its bits represent all possible binary values overn elementsLLC-Lemma: (L) = +/2 (par[,k]) > 42Long codeLXY is NP-hard

  • Long-CodeEncoding an element e[n] :Ee legally-encodes an element e if Ee = feFFTTT

  • Long-Code over Range RBP[R] the set of all subsets of R of size |R|

    Our long-code: in our context therere two types of domains R: Rx and Ry . Def: an R-long-code has 1 bit for each F P[R] namely, any Boolean function: P[R] {-1, 1}Def: a legal-long-code-word of an element eR, is a long-code ERe: P[R] {-1, 1} that assigns eF to every subset F P[R]|BP[R]| = 2|R|-1-1

  • Linearity of a Legal-EncodingAn assignment A : BP[R] {-1,1}, if legal, is a linear-function, i.e., F, G BP[R]: f(F) f(G) f(FG)

    Unfortunately, any character is linear as well!

  • The Variables of LConsider (xy) for large constant k (to be fixed later) L has 2 types of variables:a variable z[y,F] for every variable yY and a subset F BP[Ry]a variable z[x,F] for every variable xX and a subset F BP[Rx]LLC-Lemma: (L) = +/2 (par[,k]) > 42Long codeLXY is NP-hard

  • The Distribution Def: denote by the distribution over all subset of Rx, which assigns probability to a subset H as follows: Independently, for each a Rx, let aH with probability 1- aH with probability One should think of as a multiset of subsets in which every subset H appears with the appropriate probability

  • Linear equationLs multiplicative-equations are the union, over all xy , of the following: FP[RY], GP[RX] and H(RX) z(y, F) z(x, G) = z(x, F*GH)

  • Revised RepresentationMultiplicative Representation:True -1False 1L:z[X,*], z[Y,*] {-1, 1}z[X,f] z[Y, g] z[X,fgh] = 1 Representation by Fourier BasisClaim 2(par[,k]) > 42Claim 1Claim 3:The expected success of the distributional assignment on [C,V]par[,k] is at least 4 2General Fourier Analysis factsMultiplicative representation

  • Prop: if () = 1 then (L) = 1-Proof: Let A be a satisfying assignment to . Assign all variables of L according to the legal encoding of As values. A linear equation of L, corresponding to X,Y,F,G,H, would be unsatisfied exactly if xH, which occurs with probability over the choice of H. LLC-Lemma: (L) = +/2 () > 42 = 2(L) -1Note: independent of k! (Later we use that fact to define k large enough for our needs).LLC-Lemma: (L) = +/2 (par[,k]) > 42L XY is NP-hard

  • Hardness of approximating Max-E3-Lin-2Main Theorem:For any constant >0:gap-Max-E3-Lin-2(1-,+) is NP-hard.Proof:Let be a gap-3SAT-7(1, 1-) By proposition () = 1 (L) 1-

  • Lemma Main TheoremProp: Let be a constant >0 s.t.: (1-)/(+/2) 2- Let k be large enough s.t.: 43 > ()ck Then () < 1 (L) +/2 + Proof: Assume, by way of contradiction, that (L) +/2 then: 43 > ()ck () > 42,which implies that > . Contradiction!of the parallel repetition lemma

  • Long-Code as an inner product spaceDef: { A : BP[R] {-1,1} }is an inner-product space: A , B { A : BP[R] {-1,1} }

  • An Assignment to LFor any variable xX The set z[x,*] of variables of L represent the long-code of x Let be the Fourier-Coefficient

    For any variable yY The set z[y,*] of variables of L represent the long-code of y Let be the Fourier-Coefficient

  • The Distributional Assignment.Def: Let be a distributional-assignment to as follows:

    For any variable xChoose a set SRx with probability , Uniformly choose a random assignment aS.

    For any variable yChoose a set SRy with probability , Uniformly choose a random assignment bS.

  • Longcode and Fourier CoeficientsAuxiliary Lemmas: 1. For any F,GBP[R] and S R, S(FG) = S(F)S(G).

    2. For any FBP[R] and s,s R, s(F)s(F) = ss(F)

    3. For any random F (uniformly chosen) and S, E[ s(F) ]=0 and E[ (F) ]=1.=xf(x)apply multiplications commutative & associative properties(f)(f)=xf(x)xf(x)=xf(x)2x(f)=1x(f)xs, f(x) is 1 or -1 with probability go to claim2

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