1

Graph Powering Cont.

PCP proof by Irit DinurPresented by Israel Gerbi

2

Goal Reduction Goal: Input: Constraint graph (G=(V,E),C, )

where G is an (n,d, ) expander, and

< d, Output: A new graph (G’,C’) with larger gap

(denoted gap’), where

0

0

' 1min( , )

(1)

gap tgap

O t

If gap = 0,

Otherwise.

3

Last Lecture G’ Construction V’ = V B=C·t C = const. Bddd ..1 2

'

4

E’: How to generate 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. Throw edge if above path has

length>B We get weighted edges, why?

5

C’: New constraints

1 2

1 2

1 2

, '

( , ) ' ,

( '( ), '( ))

Let

C if u v such that u v is an

edge on the path from a to b

C u v

6

Plurality Assignment

Goal: We show that for every assignment ' given by

adversary, ' doesn't satisfy at least gap' edges, where

1' min( , )

(1)

tgap gap

O t

Definition: : V E is defined as follows(Informally):

( ) is the most probable value for v derived from a random walkv

Definition: ' : V' E' is defined as follows:

'( ) the opinion of w about v.vw

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New Plurality: Formal Definition

To define (v): consider the probability

distribution on vertices as follows:w V

: '( )

( ) [ ]vw w i

P i P to get from w to v

(v) is the value i that maximizes P(i)

8

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)

9

Gap’ Analysis Reminder

' [ ( , ) . '( ) ( ) '( ) ( ) ]v ugap P v u F s t a v b u

[ ( , ) . '( ) ( ) '( ) ( )] [ ( , ) ]v uP v u F s t a v b u P v u F

2

1[ ( , ) ]

2 | |P v u F

Lemma from last lesson

*

'

We want to show that [ 0] is large.

Hence gap is large.

FP N

Over all paths from a to b (weighted e’

edges)

*2

1[ 0]

2 | | FP N

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To Work… Starting with more Definitions

S := Total number of steps in our RW NF := Number of steps that were in F NF* := Number of steps that were in F,

if our RW wasn’t limited to B steps

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Motivation

In more detail, we show:

*: [ 0]FWe want to show P N is large*

FTo find P[N >0], we first find

F 2* 3 | |

(1) E[N ]8 | | |E|

t F

2F* | |

(2) E[(N ) ] O(1)t| |

F

E

F*E[N ]

2F*E[(N ) ]

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Motivation cont. Second Moment Method says:

We wanted to show:

**

*

[ ]²[ 0]

[( )²]F

FF

E NP N

E N

F 2* 3 | |

(1) E[N ]8 | | |E|

t F

2

F* | |

(2) E[(N ) ] O(1)t| |

F

E

2

2*

By using Second Moment Method we get:

3 | |8 | | |E| | |

[ 0] = | | (1) | |O(1)t| |

F

t F

t FP N

F O EE

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Back to the Beginning

'2

1[ ( , ) ]

2 | |We started with gap P weuse v u F

*

:

| |[ ( , ) ] [ 0]

(1) | |F

We now saw

t FP weuse v u F P N

O E

'2

:

1 | | | | 1min( , )

2 | | (1) | | (1) | | (1)

By combining them we get

t F t F tgap gap

O E O E O t

We can now choose t so the new gap would be twice as large!

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Expectation of NF

F(u,v) F

E[N ] = E[ ( ) ]u v F

t |F|

2 |E|

(u,v) F

[ ( ) ]E u v F

(u,v) F

t

2|E|

The graph is d regular

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Cutting Off the Tail

We will now bound

F F FS B S B

F F S B

*E[N ]= E[N ]= E[N (1- )]=

E[N ]-E[N ]

F S B

| |E[N ]

2 |E|

t F

F S BE[N ]

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The Tail

P[S>B]

F S BE[N ]F= P[S>B] E[N |S>B]

11-

B

t

(10ln| |)1

1-t

t

(10ln| |)

et

t

10

1=

| |

FE[N |S>B]| |

=E[S|S>B]| |

F

E | |

(B+t)| |

F

E | |

(20 ln | |)| |

Ft

E

2

| |

8 | | | |

t F

E

F S BE[N ]10

1 | |(20ln | |)

| | | |

Ft

E

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(1) Proof

Combining the two results above we get:

We finished proving (1). We now turn to (2)

*

F F S>B

| |E[N ] E[N ]

2 |E|

t F

2

| | | |

2 |E| 8 | | |E|

t F t F

2

2

| |(4 | | 1)

8 | | |E|

t F

2

| |(4 1)

8 | | |E|

t F

2

3 | |

8 | | |E|

t F

F 2* 3 | |

We wanted(1): E[N ]8 | | |E|

t F

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(2) Proof We now show(2): Obviously,

* 2F

| |E[(N ) ] O(1)t

| |

F

E

* 2 2F FE[(N ) ] E[(N ) ]

thF i i i step is in Fi=1

We now express N = X , where X I

2F i j i j

i,j=1 i,j=1

E[(N ) ] = E[X X ] P[X =1 X =1]

i j ii=1 j i

2 P[X =1] P[X =1|X =1]

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(2) Proof Cont.

j iP[X =1|X =1]=1 if i=j. Otherwise:

j i

th

P[X =1|X =1]= P[the walk takes at least j-i steps]

P[a walk starting from a random F endpoint takes its (j-i) step in F]

Lemma from first lesson

j-i j-i-11 | |

1- [ + ]| |

F

t E d

j ij i

P[X =1|X =1]

l l-1

l=1

1 | |= 1+ 1- [ + ]

| |

F

t E d

l l-1

l=1 l=1

| | 11+ 1- +

| |

F

E t d

| |

1+(t-1) +O(1)| |

F

E

1O(1)+ = O(1)

t

t

* 2F

| |E[(N ) ] O(1)t

| |

F

E

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(2) Proof Cont.

* 2FE[(N ) ] 2

FE[(N ) ]

i j ii=1 j i

2 P[X =1] P[X =1|X =1]

ii=1

= 2 P[X =1] O(1)

ii=1

= O(1) P[X =1]

F= O(1) E[N ]| |

= O(1) t| |

F

E

* 2F

| |E[(N ) ] O(1)t

| |

F

E

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Second Moment Method Lemma (Second Moment Method): If X is a nonnegative r.v then

Proof:

[ ]²[ 0]

[ ²]

E XP X

E X

X>0[ ²] [( )²]E X E

Cauchy Schwartz inequality

X>0[ ] [ ]E X E X

= [ ²] [X>0]E X P

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Second Moment Method Proof

We have:

Arranging:

Therefore:

[ ] [ ²] [X>0]E X E X P

[ ][X>0]

[ ²]

E XP

E X

[ ]²[ 0]

[ ²]

E XP X

E X