Inapproximability of MAX-CUT
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Inapproximability of MAX-CUT
Khot,Kindler,Mossel and O’Donnell
Moshe Ben Nehemia June 05
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Main Result
It is NP-Hard problem to approximate MAX-CUT to within a factor
is the approximation ratio achieved by the algorithm of Goemans & Williamson.
The result follows from: 1. Unique Games conjecture 2. Majority is Stablest Theorem
GW878567.0GW
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Hardness of Approximation
History: Bellare & Goldreich & Sudan :It is NP
Hard to approximate MAX-CUT within factor higher than 83/84
Hasted improved the result to 16/17 Today: closing the gap…
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Introduction
MAX-CUT: Given a Graph G =(V,E), find a partition C=(V1, V2) that
maximize:
Unique Label Cover: Given a bi-partite graph with left side vertices- V ,right side W, and edges- E each edge have a constraint bijection The goal: assign each vertex a label which satisfy the
constraint.
][][:, MMwv
)())((, vlabelwlabelwv
EVVe
ew)( 21
)(
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Unique Games Conjecture: For any there exist a constant Such that it is NP-hard to distinguish whether the
Unique Label Cover problem with label set in size M has optimum at least or at most
0, ),( MM
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Some defintions Let be an arbitrary boolean
function The influence of xi on f
Let x be a uniformly random string in :E[X]=0 and form y by flipping each bit with prob The noise stability of f for a noise rate is:
)],,,,...(),...,([Pr)( 1111}1,1{
niiinx
i xxxxxfxxffInfn
n}1,1{
)]()([Pr , yfxfyx
}1,1{}1,1{: nf
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The Correlation between x,y is define to be: E[XY] = 2 Pr[X=Y]-1
)]()([)( , yfxffS yxE
Let x be a uniformly random string in y be -correlated copy :i.e. pick each bit independently s.t. The noise correlation of f with parameter is:
n}1,1{
][ ii yxE
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Result[60’] :
arccos1))((lim 2 MajoritySn
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Fix then for any there is a small enough s.t. if is any function satisfying :
Then:
)1,0[ 00),(
}1,1{}1,1{: nf
ni
f
..1.2
0)(.1
E
)( fInfi
arccos1)( 2fS
The Majority is Stablest Theorem
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On the Geometry of MAX-CUT
The Goemans-Williamson algorithm: Embedding the graph in the unit sphere of Rn : The embedding is selected s.t. this sum is
maximize
A cut in G is obtained by choosing a random
hyperplane through the origin . And this sum bounds from above the size of the
maximal cut
Evu
vu xx),(
21
21 ,
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On the Geometry of MAX-CUT
The probability that vertics u,v lie on opposite sides of the cut is:
So the expected weight is
),arccos( ji vv
ji
jiji vvwWE ),arccos(1
][ ,
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On the Geometry of MAX-CUT
So to get:
Set the approximation ratio to:
})(arccos
{21
21
]1,1[min
GW
OPTWE ][
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Reminder
The Long Code: The codeword encoding the message is by the truth table of the “dictator”
function:
][ni
}{:)( iXiLC
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Technical Background
The Bonami Beckner operator
Proposition: Let and then:
SS
S SffT )(ˆ)(
R nf }1,1{: ]1,1[
][
2)(ˆ)(,)(nS
S sffTffS
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Technical Background
Proposition:Let then for every
Proof:Define:
And :
And using the Parseval identity we get the proposition
R nf }1,1{: ][ni
Si
i SffInf 2)(ˆ)(
2
22)()( )()()( xffInfxf iiixfxf
i
SiS
Si Sfxf:
)(ˆ)(
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Technical Background
Let and let
The k-degree influence of coordinate i on f is defined by:
Proposition: The “Majority is Stablest” Theorem remains true if we change the assumption to ')( fInf k
i
R nf }1,1{: ][ni
kSSi
ki SffInf 2)(ˆ)(
)( fInfi
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Reverse version of the “Majority is Stablest” Fix then for any there is a small enough s.t. if is any function satisfying :
Then:
]0,1( 00),(
}1,1{}1,1{: nf
ni ..1 )( fInf ki
arccos1)( 2fS
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Reverse version of the “Majority is Stablest”
Proof: Take such f, and define:
Now g holds:
And now apply the original Theorem
SoddS
Sfxfxf
xg
)(ˆ2
))()(()(
)()()(
)()(.2
0)(.1
gSgSfS
fInfgInf
gE
ii
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Reduction from Unique LC to MAX-CUT
Notations: denote the string and xy the coordinatewise product of x and y
x ),...,( )()1( nxx
Lemma 1: Completeness If ULC have OPT then MAX-CUT have
cut Lemma 2: Soundness If ULC have OPT then MAX-CUT have cut
at most
1c 212
1
s /)(arccos
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Reduction from Unique LC to MAX-CUT
Unique Label Cover
W V
'
vW’
w
MAX-CUT
j
J’ i
-{1,1}M
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Reduction from Unique LC to MAX-CUT
The Reduction: Pick a vertex at random and 2 of its
neighbors: Let and be the constrains for
those edges Let f,g be the supposed Long Codes of the labels Pick at random Pick by choosing each coordinate
independently to be 1 with probability and -1 with prob. Edge in Cut iff
VvWww ',
wv, ',' wv
Mx }1,1{M}1,1{
2121
2121
))'(()( xgxf
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Reduction from Unique LC to MAX-CUT
Completeness Assume that the LC instance has a
labeling which satisfies fraction of the edges.
now encode these labels via Long Code with prob both the edges are
satisfied by the labeling Denote the label of v,w,w’ by i,j,j’
1
21
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Reduction from Unique LC to MAX-CUT
Completeness note that: Now f,g are the Long Codes of j,j’, so: The two bits are unequal iff and that happens with prob.
hence the completeness :
)'(')( jij
1' j
'')'('
)(
))'((
)(
jijj
ij
xxxg
xxxf
2121
)()21( 21
21
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Reduction from Unique LC to MAX-CUT
Soundness – The Proof Strategy if the max-cut bigger than we’ll
be able to “decode” the “Long Code“ and create a labeling which satisfy significant fraction of the edges in the LC problem, and get a contradiction by choosing small enough.
/arccos
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Reduction from Unique LC to MAX-CUT
)]()()()(ˆ)(ˆ[
)()(
)]()'()()(ˆ)(ˆ[
')'('][',
)(,2
121
)(
'][',
',2
121
SSMSS
Sx
SS
SMSS
SSx
xxSgSfE
xx
xxSgSfE
)])'(()([]Pr[ 21
21
,,',,
xgxfaccept Exwwv
From the Fourier Transform:
)()'( )'('' xx SS
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Reduction from Unique LC to MAX-CUT
The expectation over x vanishes unless and then s,s’ have the same size. Because:
We got:
Because of for at least v in V (“good” v) We have
)'(')( ss
)(ˆ)(ˆ
)]([
)'(')(:',21
21
''
SgSf
E
SSSS
S
SSS
/arccos]Pr[accept2/
21
'
121
21 /)(arccos)]'('(ˆ[)]((ˆ[ SgESfE
wwS
S
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Reduction from Unique LC to MAX-CUT
)(arccos1)])((ˆ[(
/)(arccos)])((ˆ[(
221
221
21
21
SfE
SfE
wS
S
wS
S
S
S
w
w
M
Sh
SfESh
xfExh
h
arccos1)(ˆ
)]((ˆ[)(ˆ
)]([)(
]1,1[}1,1{:
22
1
Define
Now:
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Reduction from Unique LC to MAX-CUT
Now ,from the “Majority is stablest” theorem: We conclude that h has at least one coordinate j s.t. label the vertex v with j
)(hInf kj
)]([]))((ˆ[))]((ˆ[)(ˆ)(
212121 fInfESfESfESh k
jw
kSSj
kSSj
ww
kSSj
And we have:
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Reduction from Unique LC to MAX-CUT
From the above equation we have that for at least fraction of neighbors w of v we have
Define
And so,
Because we got that
2/)()(1
fInf k
j
}2/)(:][{][ fInfMiwCand ki
kfInfi
ki )( /2][ kwCand
2
][)(1 wCandj
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Reduction from Unique LC to MAX-CUT
Now ,if we label each vertex w in W by random element from Cand[w], then among the “good” vertices v
at least satisfied. or among the edges , and that yields the contradiction
k22
k8
2