# Graph Powering Cont.

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21-Jan-2016Category

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### Transcript of Graph Powering Cont.

Graph Powering Cont.PCP proof by Irit DinurPresentation by: Alon Vekker

Last lecture G constructionV = V B=Ct C = const.E = For every two vertices at distance at most t we have a new edge between them.

From last lectureC(u,v)VV = Vgap gap t/O(1)*min(gap,1/t)

We built the graph linearly.We look at vertices at distance at most t.We look at opinions at distance at most B.

Plurality assignmentDefinition: : 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 leastUnweighted edges!!!

Example for aWe use here : t 2 = {1,2,3}

(a): Flat pluralitya

Last week AnalysisDefinition: F is a subset of E which includes all edges that are not satisfied by .

|F|/|E|gap

We throw edges from F until |F|/|E|=min(gap,1/t)=:

Last week Analysis[e passes through F][e completely misses F]( by the lemma )( since for )

ExampleFeba12uvaToo long

Another lookIs it working?Un weighted plurality

E: What weight to give to an edge?Pick a random vertex aTake 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.

ExampleA is the plurality but they are too far.aaaaaaaaabbbabbbvbBau

Why do we get weighted edges?1311111112223323322abb

Edge Weight:(a,b) GDist(a,b) tThe weight on the adge (a,b) is:

New pluralityTo define (v): consider the probabilitydistribution on vertices as follows:Do SW starting from v, ending on w.

Lemma 1:if a path a b in G uses an edge (u,v)Then, if:(u,v) F THEN : violates the constraint on edge e.That leads us to a conclusion

When the length of the path < B

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.

Final AnalysisLemma 1Lemma 2[e completely misses F]

Proof of lemma 1:Suppose we dont stop SW after B steps Our will depend on the number of vertices.Its to big so we must stop after B steps.

calculationsLets count the probability of a path longer then B:

And therefore we get: