Tight Hardness Results for Some Approximation Problems [mostly H

1
Tight Hardness Results forSome Approximation
Problemsmostly Håstad

• Adi AkaviaDana MoshkovitzS. Safra

2
Road-Map for Chapter I
Gap-3-SAT ?
expander
Gap-3-SAT-7 ?
Parallel repetition lemma
par?, k ?
?X?Y is NP-hard
3
Maximum Satisfaction

• Def Max-SAT
• Instance
• 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 ?

4
Max-E3-Lin-2

• Def Max-E3-Lin-2
• Instance a system of linear equations L E1,
  , En over Z2each equation of exactly 3
  variables(whose sum is required to equal either
  0 or 1)
• Problem Compute ?(L)

5
Example
x1x2x31 (mod 2) x4x5x61 x7x8x91 x1x4x7
0 x2x5x80 x3x6x90

• Assigning x1-61, x7-90 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.

6
2-Variables Functional SAT

• Def ?X?Y ? over
• variables X,Y of range Rx,Ry respectively
• each ??? is of the form ?x?y Rx?Ry
• an assignment A(x?Rx,y?Ry) satisfies ?x?y
  iff ?x?y(A(x))A(y)
• Namely, every value to X determines exactly 1
  satisfying value for Y
• Thm distinguishing between
• ? A satisfies all ?
• ? A satisfies lt? fraction of ?
• IS NP-HARD as long as Rx,Rygt ?-0(1)

7
Proof Outline
Gap-3-SAT
?
Gap-3-SAT-7

• Def 3SAT is SAT where every ?i isa disjunction
  of ?3 literals.
• Def gap-3SAT-7 is gap-3SAT with theadditional
  restriction, that everyvariable appears in
  exactly 7 local-tests
• Theorem gap-3SAT-7 is NP-hard

par?, k
?X?Y is NP-hard
8
Expanders
Gap-3-SAT
?
Expanders
Gap-3-SAT-7

• Def a graph G(V,E) is a
• c-expander if for every S?V,
• S? ½V N(S)\S ? cS
• 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 cgt0
• Corollary a cut between S and V\S, for S ?
  ½V must contain gt cS edges

par?, k
?X?Y is NP-hard
9
Reduction Using Expanders

• Assume ? 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 y
• For 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 ?
  ?

constructible by the Lemma
ensuring equality
10
Correctness of the Reduction

• ? is completely satisfiable iff ? is
• In case ? is unsatisfiable ?(?) lt 1-20?/cLet
  A be an optimal assignment to ?Let Amaj assign
  xy,i the value assigned by A to the majority,
  over j, of variables xy,jLet FA and FAmaj be the
  sets of ? ? ?unsatisfied by A and Amaj
  respectively
• ?(1-?(?)) FA FA?FAmajFA\FAmaj
  ?FA?FAmaj½cFAmaj\FA ? ½cFAmaj
• and since Amaj is in fact an assignment to
  ? ?(?) ? 1- ½c(1- ?(?))/10 lt 1- ½c(20?/c)/10
  1-?

11
Notations

• Def For a 3SAT formula ? over Boolean variables
  Z,
• Let Zk be the set of all k-sequences of ?s
  variables
• Let ?k be the set of all k-sequences of ?s
  clauses
• Def For any V?Yk and C??k, 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 V?Zk, and a set
  of k clauses C??k,
• denote V C ? V is a choice of one variable of
  each clause in C.

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Parallel SAT

• Def for a 3SAT formula ? over Boolean variables
  Z, define par?, k
• par?, k has two types of variables
• yV for every set V?Yk,where yVs range is the
  set RY of all assignments to V
• xC for every set C??k,where xCs range is the
  set RX of all satisfying assignments to all
  clauses in C
• par?, 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 TC
  and TV are consistent)

RY2k
RX7k
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Gap Increases with k
Gap-3-SAT
Gap-3-SAT-7
Parallel repetition lemma

• Note that if ?(?) 1 then
• ?(par?, k) 1
• On the other hand, if ? is not satisfiable
• Lemma ?(par?, k) ? ?(?)ck for some cgt0
• Proof
• first note that 1-?(par?, 1) ? (1-?(?))/3
• now, to prove the lemma, apply the
  Parallel-Repetition lemma Raz to par?, 1

par?, k
?X?Y is NP-hard
In any assignment to ?s variables, any
unsatisfied clause in ? induces at least 1 (out
of corresponding 3) unsatisfied ?? par?, 1
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Conclusion ?X?Y is NP-hard

• Denote
• ? par?, k
• XxC
• YyV
• Then,
• distinguish between
• ? A satisfies all ?
• ? A satisfies lt? fraction of ?
• IS NP-HARD as long as Rx,Rygt ?-0(1)

15
Road-Map for Chapter II
?X?Y is NP-hard
Long code
L? ?
?(?) ??(?) ?
LLC-Lemma? ?(L?) ½?/2 ? ?(par?,k) gt 4??2
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Main Theorem

• Thm gap-Max-E3-Lin-2(1-?, ½?) is NP-hard.
• That is, for every constant 0lt?lt¼ it is NP-hard
  to distinguish between the case where ?1-? of the
  equations are satisfiable and the case where ?½?
  are.
• It is therefore NP-Hard to approximateMax-E3-Li
  n-2 to within factor 2-? for any constant 0lt?lt¼

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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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Distributional Assignments

• Let ? be a SAT instance over variablesZ of range
  R.
• Let ?(R) be all distributions over R
• Def a distributional-assignment to ? is A Z ?
  ?(R)Denote by ??(?) themaximum over
  distributional-assignments A of theaverage
  probability for ??? to be satisfied,if
  variables values are chosen according to A
• Clearly ?(?) ? ??(?). Moreover
• Prop ?(?) ? ??(?)

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Distributional-assignment to ?
1
1
0
1
0
0
0
1
x1
x3
x2
xn
OR
1
0
1
1
0
0
1
0
x1
x3
x2
xn
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Restriction and Extension

• Def
• For any y?Y over RY and x?X over RX s.t ?x?y??
• The natural restriction of an a?RX to Ry is
  denoted ay
• The elevation of a subset F?PRY to RX is the
  subset F?PRX of all members a of RX for which
  ?x?y(a)? F F a ay ? F

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Long-Code
?X?Y is NP-hard
Long code
L?

• In the long-code the set of legal-words consists
• of all monotone dictatorships
• This is the most extensive binary code,
• as its bits represent all possible binary values
  over
• n elements

LLC-Lemma? ?(L?) ½?/2 ? ?(par?,k) gt 4??2
22
Long-Code

• Encoding an element e?n
• Ee legally-encodes an element e if Ee fe

T
F
F
T
T
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Long-Code over Range R

• BPR ? the set of all subsets of R of size
  ½R
• 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 ? PR
  namely, any Boolean function PR ? -1, 1
• Def a legal-long-code-word of an element e?R, is
  a long-code ERe PR ? -1, 1
• that assigns e?F to every subset F ? PR

BPR 2R-1-1
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Linearity of a Legal-Encoding

• An assignment A BPR ? -1,1, if legal, is a
  linear-function, i.e., ? F, G ? BPR f(F) ?
  f(G) ? f(F?G)
• Unfortunately, any character is linear as well!

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The Variables of L
?X?Y is NP-hard
Long code
L?

• Consider ? (?x?y??) forlarge constant k (to
  befixed later)
• L? has 2 types of variables
• a variable zy,F for every variable y?Y and a
  subset F ? BPRy
• a variable zx,F for every variable x?X and a
  subset F ? BPRx

LLC-Lemma? ?(L?) ½?/2 ? ?(par?,k) gt 4??2
26
The Distribution ??

• Def denote by ?? the distribution over all
  subset of Rx, which assigns probability to a
  subset H as followsIndependently, for each a ?
  Rx, let
• a?H with probability 1-?
• a?H with probability ?

One should think of ?? as a multiset of subsets
in which every subset H appears with the
appropriate probability
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Linear equation

• L?s multiplicative-equations are the union, over
  all ?x?y ? ?, of the following?F?PRY,
  G?PRX and H???(RX) z(y, F) ? z(x, G) z(x,
  F?G?H)

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Revised Representation
Multiplicative representation
General Fourier Analysis facts
Representation by Fourier Basis

• Multiplicative Representation
• True ? -1
• False ? 1
• L
• zX,, zY, ? -1, 1
• zX,f zY, g zX,fgh 1

Claim 2
Claim 1
Claim 3The expected success of the
distributional assignment ? on ?C,V?par?,k is
at least 4? ?2
?(par?,k) gt 4??2
?
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?X?Y is NP-hard
L? ?

• Prop if ?(?) 1 then ?(L?) 1-?
• ProofLet 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 x?H, which occurs with
  probability ? over the choice of H.
• LLC-Lemma ?(L?) ½?/2 ? ?(?) gt 4??2

LLC-Lemma? ?(L?) ½?/2 ? ?(par?,k) gt 4??2
Note independent of k! (Later we use that fact
to define k large enough for our needs).
? 2?(L) -1
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Hardness of approximating Max-E3-Lin-2

• Main Theorem
• For any constant ?gt0
• gap-Max-E3-Lin-2(1-?,½?) is NP-hard.
• Proof
• Let ? be a gap-3SAT-7(1, 1-?)By
  proposition ?(?) 1 ? ?(L?) ? 1- ?

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Lemma ? Main Theorem

• PropLet ? be a constant gt0 s.t. (1-?)/(½?/2)
  ? 2-?Let k be large enough s.t. 4?3 gt
  ?(?)ckThen ?(?) lt 1 ? ?(L?) ? ½?/2 ? ½ ?
• Proof Assume, by way of contradiction, that ?(L)
  ? ½?/2 then4?3 gt ?(?)ck ? ?(?) gt 4??2,
• which implies that ? gt ?. Contradiction!

of the parallel repetition lemma
32
Long-Code as an inner product space

• Def
• A BPR ? -1,1
• is an inner-product space
• ? A , B ? A BPR ? -1,1

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An Assignment ? to L?

• For any variable x?XThe set zx, of variables
  of L? represent the long-code of x Let
  be the Fourier-Coefficient lt?zX,,?sgt
• For any variable y?YThe set zy, of variables
  of L? represent the long-code of y Let
  be the Fourier-Coefficient lt?zY,,?sgt

34
The Distributional Assignment.

• Def
• Let ? be a distributional-assignment to ? as
  follows
• For any variable x
• Choose a set S?Rx with probability ,
• Uniformly choose a random assignment a?S.
• For any variable y
• Choose a set S?Ry with probability ,
• Uniformly choose a random assignment b?S.

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Longcode and Fourier Coeficients
go to claim2

• Auxiliary Lemmas
• 1. For any F,G?BPR and S ? R,
• ?S(FG) ?S(F)?S(G).
• 2. For any F?BPR and s,s ? R,
• ?s(F)?s(F) ?s?s(F)
• 3. For any random F (uniformly chosen) and S??,
• E ?s(F) 0 and E ??(F) 1.

???x??f(x) apply multiplications commutative
associative properties
??(f)??(f)?x??f(x)?x??f(x) ?x????f(x)2?x????(
f)1?x????(f)
?x?s, f(x) is 1 or -1 with probability ½
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Home Assignment

• Given an assignment to a Longcode ABPR ? -1,
  1, show that for any (constant) ? gt 0, there is
  a constant h(?), which depends on ?, however does
  not depend on R such that e ? R ?(Ee, A)
  gt ½ ? ? h(?)where ?(A1, A2) is the
  fraction of bits A1 and A2 differ on.

37
Whats Ahead

• We show ?s expected success on ?x?y?? is gt 4??2
  in two steps
• First we show (claim 1) that ?s success
  probability, for any ?x?y?? is
• Then show (claim 3) that value to be ? 4??2

38
Claim 1
General Fourier Analysis facts
Multiplicative representation
go to claim3
Representation by Fourier Basis

• Claim 1
• The success probability of ? on
• ?x?y ?? is
• ProofThat success probability is at least
• and if SSy there is at least one b?S s.t. by
  ? S
• So, ?s success probability is at least S-1
  times the case in which the chosen S and S
  satisfy Sy S,
• i.e. at least

Claim 1
Claim 2
Claim 3The expected success of the
distributional assignment ? on ?C,V?par?,k is
at least 4? ?2
?(par?,k) gt 4??2
?
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Lemmas Proof - Claim 2 (1)
go to claim3
General Fourier Analysis facts
Multiplicative representation

• Claim 2
• Proof
• The test accepts iff zy,Fzx, Gzx,FGH
  1
• By our assumption, this happens with probability
  ?/2½.
• Now, according to the definition of the
  expectation
• E?x?y, F, G, Hzy,Fzx, Gzx,FGH
• 1(½?/2) (-1)(1 -(½?/2)) ?

Representation by Fourier Basis
Claim 2
Claim 1
Claim 3The expected success of the
distributional assignment ? on ?C,V?par?,k is
at least 4? ?2
?(par?,k) gt 4??2
?
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Lemmas Proof - Claim 2 (2)

• We next show that
• Hence,

41
Lemmas Proof - Proposition
42
Lemmas Proof - Claim 3

• Claim 3 The expected success of the
  distributional assignment ? on ?x?y ?? is at
  least 4??2
• Proof Claim 1 gives us the initial lower bound
  for the expected success

43
Lemmas Proof - Claim 3

• As weve already seen, . Hence, our
  lower-bound takes the form of
• Or alternatively,
• Which allows us to use the known inequality
  Ex2?Ex2 and get

44
Lemmas Proof - Claim 3

• By auxiliary lemmas (4??)-1/2 ? e-2?? ?
  (1-2?)?, i.e. ?-1/2 ?(4?)1/2 (1-2?)?,
  which yields the following bound
• That is,
• Now applying claim 2 results the desired lower
  bound

45
Lemmas Proof -Conclusion

• We showed that there is an
• assignment scheme with expected
• success of at least 4??2 ,
• ? There exists an assignment that satisfies at
  least 4??2 of the tests in ?
• ? ?(?) gt 4??2
• Q.E.D.

46
Home Assignment

• Show it is NP-hard, for any ? gt 0, given a 3SAT
  instance ?, to distinguish between the case where
  ?(?) 1, and the case in which ?(?) lt 7/8?
• Hint Let ?s variables be as in L?, and ?s
  clauses to take the form F OR G OR F?G?Hfor
  f and g chosen in the same way as in L?,while h
  is chosen as follows
• H(b) 1 for b such that F(bV) and G(b) are both
  FALSE
• For all other bs, independently for each b,
  H(b)1 with probability ?, and -1 with
  probability 1-?