Graph Powering Cont'

1
Graph Powering Cont.

• PCP proof by Irit Dinur
• Presentation by Alon Vekker

2
Last lecture G construction

• V V
• BCt C const.
• E For every two vertices at distance at most t
  we have a new edge between them.

3
From last lecture
V
V V
S
C(u,v)
gap t/O(1)min(gap,1/t)
gap
4
We built the graph linearly.

• We look at vertices at distance at most t.
• We look at opinions at distance at most B.

5
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!!!

6
Example for s
We use here t 2 S 1,2,3
a
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s(a) Flat plurality
a
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Last week Analysis

• Definition F is a subset of E which includes all
  edges that are not satisfied by s.
• F/Egap
• We throw edges from F until F/Emin(gap,1/t)
  

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Last week Analysis
e passes through F
e completely misses F
( by the lemma )
( since for
)
10
Example
Too long
a
e
2
a
b
1
v
u
F
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Another look
Is it working?
Un weighted plurality
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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.

13
Example

• A is the plurality but they are too far.

B
a
a
b
a
a
u
b
a
b
a
a
b
v
b
b
a
a
b
a
a
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Why do we get weighted edges?
2
3
2
b
3
2
2
a
1
2
3
3
1
3
2
1
1
b
1
1
1
1
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Edge Weight

• (a,b) G
• Dist(a,b) t
• The weight on the adge (a,b) is

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New plurality

• To define (v) consider the probability
• distribution on vertices as follows
• Do SW starting from v, ending on w.

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Lemma 1

• if a path a b in G uses an edge (u,v)
• Then, if
• (u,v) F
• THEN s violates the constraint on edge e.
• That leads us to a conclusion
• When the length of the path lt B

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Lemma 2

• Let G be an (n,d,?)-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.

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Final Analysis
e completely misses F
Lemma 1
Lemma 2
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Proof of lemma 1

• Suppose we dont stop SW after B steps
• Our S will depend on the number of vertices.
• Its to big so we must stop after B steps.

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calculations

• Lets count the probability of a path longer then
  B
• And therefore we get