1 Graph Powering Cont. PCP proof by Irit Dinur Presented by Israel Gerbi
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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
7
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
10
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
11
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 ) ]
12
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
13
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!
14
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
15
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 ]
16
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
17
(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]
19
(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
20
(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
21
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
22
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
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