Transcript Tight Hardness Results for Some Approximation Problems [Raz,Håstad,...] Adi Akavia Dana Moshkovitz & Ricky Rosen S.

Tight Hardness Results for
Some Approximation
Problems
[Raz,Håstad,...]
Adi Akavia
Dana Moshkovitz & Ricky Rosen
S. Safra
1
“Road-Map” for Chapter I
Gap-3-SAT 
expander
Gap-3-SAT-7 
Parallel repetition
lemma
par[, k] 
XY is NP-hard
2
Maximum Satisfaction
Def: Max-SAT

Instance:



Maximization:


A set of variables Y = { Y1, …, Ym }
A set of Boolean-functions (local-tests) over Y
 = { 1, …, l }
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, Y’s range and 
3
Max-E3-Lin-2
Def: 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)
4
Example
x1+x2+x3=1 (mod 2)
x4+x5+x6=1
 Assigning x1-6=1, x7-9=0
x7+x8+x9=1
satisfy all but the third
x1+x4+x7=0
equation.
x2+x5+x8=0
 No assignment can satisfy
x3+x6+x9=0
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.
5
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 < fraction of 
is NP-hard as long as |Rx|,|Ry|> -0(1)
6
Proof Outline
?
Def: 3SAT is SAT where every i is
a disjunction of 3 literals.
Gap-3-SAT
Gap-3-SAT-7
par[, k]
XY is NP-hard
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
7
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| ½|V|: |N(S)\S|  c·|S|
par[, k]
XY is NP-hard
[where N(S) denotes the set of neighbors of S]
Lemma: For every m, one can construct in polytime a 3-regular, m-vertices,
c-expander, for some constant c>0
Corollary: a cut between S and V\S, for |S| 
½|V| must contain > c·|S| edges
8
Reduction Using Expanders
Assume ’ for which (’) is either 1 or 120/c.
 is ’ with the following changes:
•
an occurrence of y in i is replaced by a
variable xy,i
constructible by the Lemma
•
•
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:
1.
2.
ensuring equality
||  10 |’|
Each variable xy,i of  appears in exactly 7 i 

9
Correctness of the Reduction
1.
2.
 is completely satisfiable iff ’ is
In 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| = |FAFAmaj|+|FA\FAmaj| 
|FAFAmaj|+½c|FAmaj\FA|  ½c|FAmaj|
and since Amaj is in fact an assignment to ’
()  1- ½c(1- (’))/10 < 1- ½c(20/c)/10= 1-
10
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: Let Y=Zk and X=k, and

RY be the set of all assignments to k variables

RX be the set of all sat. assignments to k clauses
Def: For any set of k variables yZk, and a set of k
clauses xk,
denote y x  y is a choice of one variable of each
clause in x.
11
Parallel SAT
Def: for a 3SAT formula  over Boolean
variables Z, let par[, k] be the following
2-var functional SAT
par[, k]’s variables:


Y=Zk (one var. for each k z’s) each y‘s range
is RY
|RY|=2k
Xk (one var. for each k ‘s) each x‘s range
is RX
|RX|=7k
par[, k] local-tests: xy for every y x
which accepts if x’s value restricted to y
is y’s value
[namely, if the assignments are consistent]
12
Gap Increases with k
Note that if () = 1 then
(par[, k]) = 1
Parallel
repetition
lemma
Gap-3-SAT
Gap-3-SAT-7
par[, k]
XY is NP-hard
On the other hand, if  is not satisfiable:
Lemma: (par[, k])  ()c·k for some c>0
In any assignment to s variables, any
unsatisfied clause in  ”induces“ at least 1 (out
of corresponding 3) unsatisfied  par[, 1]
Proof:
first note that
1-(par[, 1])  (1-())/3
now, to prove the lemma, apply the ParallelRepetition lemma [Raz] to par[, 1]
13
Conclusion: XY is NP-hard

distinguish between:
 A satisfies all 
  A satisfies <p fraction of 
is NP-hard as long as |Rx|,|Ry| p-0(1)

14
“Road-Map” for Chapter II

XY
is NP-hard
Long code
L 
() = () 
LLC-Lemma:
(L) = ½+/2  (par[,k]) > 42
15
Main Theorem
Thm: gap-Max-E3-Lin-2(1-, ½+) is NPhard.
That is, for every constant 0<<¼ it is NPhard to distinguish between the case
where 1- of the equations are
satisfiable and the case where ½+ are.
[ It is therefore NP-Hard to approximate
Max-E3-Lin-2 to within factor 2- for any
constant 0<<¼]
16
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).
17
Distributional
Assignments
Long
code
Let  be a SAT instance over variables
Z of range R.
Let (R) be all distributions over R
XY is
NP-hard
L 
LLC-Lemma:
(L) = ½+/2 
(par[,k]) > 42
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: ()  ()
18
Distributional-assignment to 
1
0
1
1
0
x
1
0
0
1
x
2
x
3
x
n
19
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 a|y
 The elevation of a subset FP[RY] to RX
is the subset F*P[RX] of all members a
of RX for which xy(a) F
F* = { a | a|y  F }
20
Long-Code
Long
code
XY is
NP-hard
L
In the long-code the set of legal-words consists
of all monotone dictatorships
LLC-Lemma:
This is the most extensive binary code,
(L) = ½+/2 
(par[,k]) > 42
as its bits represent all possible binary values over
n elements
21
Long-Code


Encoding an element e[n] :
Ee legally-encodes an element e if Ee = fe
F
F
T
T
T
22
Long-Code over Range R
BP[R]  the set of all
subsets of R
of size ≤½|R|

|BP[R]| = 2|R|-1-1
Our long-code: in our context there’re two
types of domains “R”: Rx and Ry .
Def: an R-long-code has 1 bit for each F  BP[R]
Def: a legal-long-code-word encoding an element
eR, is ERe: BP[R]  {-1, 1} that assigns the
Boolean value eF to every subset F  BP[R]
23
Linearity of a Legal-Encoding
An assignment A : BP[R]  {-1,1}, if legal,
is a linear-function, i.e.,  F, G  BP[R]:
A(F)  A(G)  A(FG)
Unfortunately, any character is linear as
well
24
The Variables of L
Consider  (xy) with
small constant p (to be
fixed later)
Long
code
XY is
NP-hard
L
LLC-Lemma:
(L) = ½+/2 
(par[,k]) > 42
L has 2 types of variables:
1.
2.
a variable z[y,F] for every variable y
of  and a subset F  BP[RY]
a variable z[x,F] for every variable x
of  and a subset F  BP[RX]
25
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


aH with probability 
aH with probability 1-
One should think of  as a multiset of subsets in which
every subset H appears with the appropriate probability
26
Linear equation
L‘s linear-equations are the union,
over all xy  par[,k], of:
 xy , F, G, H there’s a test
z[y, F] + z[x, G] = z[x, F*  G  H ] (mod 2)
‘F*GH’ is the symmetric
difference of the extension
of F to SC, G and H
27
Multiplicative
representatio
n
Revised Representation
Multiplicative Representation:
 True  -1
 False  1
 L:


General
Fourier
Analysi
s facts
Representation
by Fourier
Basis
Claim
2
Claim
1
Claim 3:The expected success of
the distributional assignment  on
[C,V]par[,k] is at least 4 2

(par[,k]) > 42
z[X,*], z[Y,*]  {-1, 1}
z[X, F] • z[Y, G] • z[X, F* • G • H ] = 1
28
XY is
NP-hard
L 
Prop: if () = 1 then (L) = 1-
LLC-Lemma:
Proof:
(L ) = ½+/2 
(par[,k]) > 42
Let A be a satisfying assignment
to . Assign all variables of L
according to the legal encoding of A’s values.
A linear equation of L, corresponding to
X,Y,F,G,H, would be unsatisfied exactly if
A(x)H, which occurs with probability  over
the choice of H. Note: independent of p! (Later we

define p small enough).
LLC-Lemma: (L) = ½+/2  () > 42
 = 2(L) -1
29
Hardness of approximating Max-E3-Lin-2
Main Theorem:
For any constant >0:
gap-Max-E3-Lin-2(1-,½+) is NPhard.
Proof:
By proposition
(’) = 1  (L’)  1- 
30
Lemma  Main Theorem
of the parallel
Prop:
repetition lemma
Let  be a constant >0 s.t.:
(1-)/(½+/2)  2-
Set p < 43
Then () < 1  (L’)  ½+/2  ½+ 
Proof: Assume, by way of contradiction,
that (L)  ½+/2 then:
43 > (’)c·k  () > 42,
which implies that  > . Contradiction!
31
An Assignment  to L
For any variable x of 
The set z[x,*] of variables of L represents the longcode of x
Let Ax (S) be the Fourier-Coefficient <|z[x,*],s>
For any variable Y of 
The set z[Y,*] of variables of L represent the longcode of y
Let Ay (S) be the Fourier-Coefficient <|z[y,*],s>
33
The Distributional Assignment
Def:
Let  be a distributional-assignment to  as follows:


For any variable x
2
 Choose a set SRx with probability  Ax (S)  ,
 Uniformly choose a random assignment a S.
For any variable y
2
Ay  S  ,
 Choose a set S Ry with probability
 Uniformly choose a random assignment b S.


34
Home Assignment

Given an assignment to a Longcode
A:BP[R]  {-1, 1}, show that for any
(constant)  > 0:
| {e  R | (Ee, A) > ½ + ½ } |  -2
where (A1, A2) is the fraction of bits
A1 and A2 differ on.
35
What’s Ahead:
We would show that ‘s expected success on
xy  is > 42 in two steps:
First we show (claim 1) that ‘s success
probability, for any xy is

2
 A S| y 
SRx
y

 Ax S
2

S
Then show (claim 3) that value to be  42
36
Multiplicative
representation
Claim 1
go to claim3
Representation
by Fourier
Basis
Claim 1:
The success probability of  on
xy  is

   A S|y
y
SRx 

Claim
2
2
 Ax S 
2

S -1 

Proof:
That success probability is at least
A S'  A S  Pr
  A  S    A  S   Pr
2

2
y
S'Ry ,SRx
x
2

SRx
y
|y
bR S
2
x
General
Fourier
Analysis
facts
bR
Claim 1
Claim 3:The expected success of
the distributional assignment  on
[C,V]par[,k] is at least 4 2

(par[,k]) > 42


b | y  S'
b | y  S|y 


S
and there is at least one bS s.t. b|y  S’
37
Lemma’s Proof - Claim 2 (1)
go to claim3
Claim 2:

2
S
  Ay S|y  Ax S  1  2ε    δ
Ε
xy SR

x

 

Proof:
The test accepts iff z[y, F]•z[x, G]•z[x,F*•G•H] = 1
By our assumption, this happens with probability /2+½.
Now, according to the definition of the expectation:
Exy, f, g, h [z[y, F]•z[x, G]•z[x, F*•G•H]] =
1•(½+/2) + (-1)•(1 -(½+/2)) = 
38
Lemma’s Proof - Proposition
EF,G,H z  y, F   z x, G   z x, F*  G  H 
 A S   A S   1  2ε
SRX
y
x
|y
|S|
2
EF,G,H z  y, F   z  x, G   z  x, F*  G  H 


A T  

Ay  T  Ax  S1  Ax  S2  EF,G,H  T F  S1  G  S2 F * G H   

Ay  T  Ax  S1  Ax  S2  EF,G,H T F  S2|y F *  S1  G  S2  G  S2 H   

Ay  T  Ax  S1  Ax  S2  EF [T
T RY ,
S1 RX ,S2 RX

T RY ,
S1 RX ,S2 RX

T RY ,
S1 SC ,S2 SC

T RY ,
S1 RX ,S2 RX

y
T
F     Ax  S1  S  G     Ax S2  S F * G H   
1
 A S  A S  A S  1  1  1  2 
SRx
y
|y
x
x
2
S2|y
S
F  ]  EG [S

1
S2
 G  ]  EH [χS H  ] 
2
 A S  A S   1  2 
S Rx
y
|y
x
2
S
39
Lemma’s 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:
Ex  y


2
2
1
  (Ay  S | y )  (Ax  S )  | S | 
SRx
40
Lemma’s Proof - Claim 3
 .1 Hence, our
As we’ve already seen, 
lower-bound takes the form of
Ax S 
s Rx
Ex  y [(
 (A S | y )
SRx
y
2
2
 (Ax S )  | S | )  (
2
1
2
 A S )]
SRx
x
Or alternatively,
E [(  A  S | y   (A S )  | S | ) ]
Rx
Which allows us to use the known
inequality E[x2]E[x]2 and get
x y
S
(Ex y [
y

R
S x
x
Ay  S | y
 12 2
2
  (A S )
x
2
 12
| S|
])2
41
Lemma’s Proof - Claim 3
By auxiliary lemmas (4|S|)-1/2  e-2|S| 
(1-2)|S|, i.e. |S|-1/2 (4)1/2 ·(1-2)|S|,
which yields the following bound
(Ex  y [

SRx
Ay  S | y   (Ax  S )2  (4)
 12
|S|
 1  2 
])2
That is,
4(Ex  y [  Ay  S | y   (Ax  S )2  1  2 
|S|
])2
S X
Now applying claim 2 results the desired
lower bound
4  (E[

SRx
Ay  S | y   (Ax  S )2  1  2  ])2  4
|S|
2
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Lemma’s 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 
 () > 42
Q.E.D.
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Home Assignment
Show it is NP-hard, for any  > 0, given a 3SAT instance
, to distinguish between the case where () = 1,
and the case in which () < 7/8+
Hint: Let ’s variables be as in L, and ’s clauses to
take the form
f OR g OR ‘f* + g + h’
for 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(b|V) and g(b) are both FALSE
For all other b’s, independently for each b, h(b)=1 with
probability , and 0 with probability 1-
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