PCP Theorem By Gap Amplification

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PCP Theorem By Gap Amplification

• Result by Irit Dinur

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INPUT x
RAND BITS 0 1 0 1 0 0 0
PROOF FOR x
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• Constraint graph Gh(V,E),?,Ci
• (V,E) is a graph
• Vµ? is also set of variables
• C(e)µ?2 is the set of constraints
• size(G)?(VE.?2)
• UNSAT(G)???

Constraint graph satisfaction is NP-hard.
Hardness of Gap-ConsGraph implies hardness of
Gap-3SAT
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• Gap-GonsGraph is NP-hard for ?lt1/n
• So we just have to worry about ?1/n
• We will use a process that almost lossless
  enlarges value ?

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preprocessing
Size(G) and ? dont change a lot
powering
composition
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• Preprocessing
• 9 ?ltd , ?1
• G ) G (conversion)
• G is d-regular with self-loops
• ?(G)?ltd
• G and G have the same alphabet
• size(G)O(size(G))
• ?1.UNSAT(G) UNSAT(G) UNSAT(G)

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Set all red edges to trivial constraints
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preprocessing
Size(G) and ? dont change a lot
powering
composition
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preprocessing
Size(G) and ? dont change a lot
powering
composition
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• Constraint Satisfaction Problem (CSP)
• F(V, ?, C)
• Vx1,x2,,xn xi2?
• CC1,C2,,Cm Ci µ ? q
• UNSAT(F) ??

• PCP Reduction Parameters
• ? probability of error
• R number of final constraints
• s size of each constraint
• q of variables in each constraint
• ? alphabet of final CSP

• PCP Reduction
• F(V,?,C) ) F(V,?,C)
• UNSAT(F)0 ) UNSAT(F)0
• UNSAT(F)? 0 ) UNSAT(F) gt?

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• Assignment Tester
• F(R, s, q, ?, ?)
• Input Boolean circuit ? of size n over Boolean
  variables X
• Output a CSP ??1,?2,,?R over variables X
  and Y
• Size of each ?i is bounded to s(n)
• Each ?i depends on q(n) variables. Variables in
  Y take values from an alphabet ?.
• For every Boolean assignment a to variables in X
  
• If a satisfies ?, then it has an extension to Y
  which satisfies ?.
• If a is ?-far from satisfying ? , then every
  extension of a makes at least ? fraction of
  constraints wrong.

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• Assignment Tester
• F(R, s, q, ?, ?)
• Input Boolean circuit ? of size n over Boolean
  variables X
• Output a CSP ??1,?2,,?R over variables X
  and Y
• Size of each ?i is bounded to s(n)
• Each ?i depends on q(n) variables. Variables in
  Y take values from an alphabet ?.
• For every Boolean assignment a to variables in X
  
• If a satisfies ?, then it has an extension to Y
  which satisfies ?.
• If a is ?-far from satisfying ? , then every
  extension of a makes at least ? fraction of
  constraints wrong.

• Assignment Tester
• F(?)
• Input Boolean circuit ? of size n over Boolean
  variables X
• Output Gh (XUY, E), ?, C i
• For every Boolean assignment a to variables in X
  
• If a satisfies ?, then it has an extension to Y
  which satisfies G.
• Otherwise for every extension b of a we have
• UNSATb(G) ? . dist(a, SAT(?))

Well!! Does such a thing exist??
YES !
In fact the original proof of PCP Theorem
generates an assignment tester.
WAIT! What kind of proof is this??
There are easier ways to construct one.
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• Composition
• If an assignment tester F with alphabet ?0 exists
  then every constraint graph G can be converted to
  G with alphabet ?0 where
• size(G)M.size(G)
• c.UNSAT(G) UNSAT(G) UNSAT(G)

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preprocessing
Size(G) and ? dont change a lot
powering
composition