Graph Powering Cont.
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1 Graph Powering Cont. PCP proof by Irit Dinur Presentation by: Alon Vekker
description
Graph Powering Cont. PCP proof by Irit Dinur Presentation by: Alon Vekker. Last lecture G’ construction. V’ = V B=C · t C = const. E’ = For every two vertices at distance at most t we have a new edge between them. From last lecture. V. V’ = V. Σ. C(u,v). gap’ ≥ t/O(1)*min(gap,1/t). - PowerPoint PPT Presentation
Transcript of Graph Powering Cont.
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Graph Powering Cont.
PCP proof by Irit DinurPresentation by: Alon Vekker
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Last lecture G’ construction V’ = V B=C·t C = const. E’ = For every two vertices at distance
at most t we have a new edge between them.
Bddd ..1 2
'
),(:)()(,),(' 2121 vvCtNtNvvbaC ba
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From last lecture
),(:)()(,),(' 2121 vvCtNtNvvbaC ba
Output ParamsInput Params
Vertex set
alphabet
Constraints
gapC(u,v)
Σ
VV’ = V
gap gap’ ≥ t/O(1)*min(gap,1/t)
Bddd ..1 2
'
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We built the graph linearly.
We look at vertices at distance at most t.
We look at opinions at distance at most B.
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Plurality assignment
Definition: : V’ is defined as follows: the opinion of w about v.
Definition: : V is defined as follows: (v) is the plurality of opinions of about v. Plurality : al least
Unweighted edges!!!
'
'
'vw)('
1
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Example for σ’
2)('1va
3)('2va
1)('4va
a
1)('6va
1)('5va 3)('
3va
We use here : t ≤ 2
Σ = {1,2,3}6v
1v
5v 4v 3v
2v
)2(aN
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σ(a): Flat plurality
2)(' 1 av
2)(' 3 av1)(' 5 av
a
1)(' 6 av
6v
1v
5v 4v 3v
2v1)(' 2 av
3)(' 4 av
1)( a
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Last week Analysis Definition: F is a subset of E which
includes all edges that are not satisfied by σ.
|F|/|E|≥gap
We throw edges from F until |F|/|E|=min(gap,1/t)=:
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Last week Analysis
[e’ passes through F]
[e’ completely misses F]
( by the lemma )
( since for )(1 ) 1t t 1/ t
2'
1'
e
gap P
2'
1(1 )
eP
2
1(1 (1 ) )t
F
E
2
1(1 (1 ))t
2,
t
12
1 | |(1 (1 ( ( ) ) )
| |t tF
E d
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Example
'( ) 1?ua '( ) 2vb
( ) 4uN B B
F
e’ba 1 2
u v
a
)),()()(')()('(' FuvubvaPgap uv
5t Too long
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Another look
)),()()(')()('(' FuvubvaPgap uv
)),(()),(|)()(')()('( FuvusewePFuvuseweubvaP uv
2
1( ( , ) )P we use v u F
Is it working?Un weighted plurality
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1 | |(1 (1 ( ( ) ) )
| |t tF
E d
!!!No
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E’: What weight to give to an edge?
Pick a random vertex a Take a step along a random edge
out of the current vertex. Decide to stop with probability 1/t. Stop if you passed B steps already.
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Example A is the plurality but they are too far.
( ) ?v
'( ) 1ua
'( ) 2vb
a
a
a
a
a
a
a
a
a
bb
ba
b
b
b
v
b
B
a
u
( ) 2v
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Why do we get weighted edges?
]'[1
]''[''2'
FthroughpassesePeviolatesePgapee
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1
1
1
1 1 1 1
22
2
3
32
3
32
2a
b
b
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Edge Weight: (a,b) G’ Dist(a,b) ≤ t The weight on the adge (a,b) is:
1w
1 1 1P (1 ) ( )l l B l length of the path
t t d
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New plurality
To define (v): consider the probabilitydistribution on vertices as follows:
Do SW starting from v, ending on w.
Vw
1)(':
)1)((vwwwPvP
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Lemma 1: if a path a b in G uses an edge (u,v) Then, if:
(u,v) F THEN : σ’ violates the constraint on edge
e’.That leads us to a conclusion…
When the length of the path < B
)()(' ua u
)()(' vb v
22
1)),(|)()(')()('(
FuvuseweubvaP uv
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Lemma 2:
Let G be an (n,d,λ)-expander and F subset of E. Then the probability that a random walk, starting in the zero-th step from a random edge in F, passes through F on its t step is bounded by
Later used to prove PCP theorem.
1tF
E d
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Final Analysis
)),()()(')()('(' FuvubvaPgap uv
)),(()),(|)()(')()('( FuvusewePFuvuseweubvaP uv
2
1( ( , ) )
2P we use v u F
Lemma 1
Lemma 2
2
1(1 (1 ) )
2tF
E
2
1(1 (1 ))
2t
2,
2
t
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1 | |(1 (1 ( ( ) ) )
2 | |t tF
E d
[e’ completely misses F]2'
1(1 )
2 eP
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Proof of lemma 1: Suppose we don’t stop SW after B
steps Our Σ will depend on the number of
vertices. Its to big so we must stop after B
steps.
2
1)),(|)()(')()('(
FuvuseweubvaP uv
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calculations Lets count the probability of a path
longer then B:
And therefore we get:
)2ln(22
1)1
1(2
Cet
CB
222 2
1
2
11)),(|)()(')()('(
FuvuseweubvaP uv
(1 ) xx e
