The PCP Theorem via gap amplification

1
The PCP Theoremvia gap amplification

• Irit Dinur
• Presentation by Michal Rosen Adi Adiv

2
The PCP Theorem

• NP PCP½log(n),O(1)
• What does it mean?

3
Definitions

• NP
• L?NP if there is a protocol ltp,vgt
• p prover has infinite running time.
• v verifier has O(poly(n)) running time.
• s.t.
• if x?L
• if x?L

?p v(p) 1
?p v(p) 0
4
Definitions

• NP-Complete
• L?NPC if L?NP and
• if for ?A?NP A p L.
• We saw that SAT, 3-Col, Ham
• Are in NPC

5
PCP
Prover gives a proof
1
0
0
1
1
0
0
1
Hide it, and gives it to the verifier Verifier
picks constant number of cells randomly Accept/rej
ect according the check
6
Definitions
The probability upon the random coins

• PCP?r,q
• L?PCP?r,q if there is a protocol ltp,vgt
• s.t. if x?L
• if x?L
• r how many coins does v need for checking p
• q the number of places v samples p

?p Prv accepts p 1
?p Prv accepts p ?
7
Example -3-col
P gives a proof (coloring) ? 0,1,2 V
choose an edge randomly and check it
2
0
0
1
u x v y
V accepts V regects
u
v
y
x
8
Example 3-col
q the number of places v samples p

• 3-Col
• Prover
• Verifier
• q
• r
• ?

gives a coloring of G.
r how many coins does v need for checking p
chooses edge (u,v) randomly, checks u and v
have different colors.
O(1)
O(log(n))
? if x ? L ?p Prv accepts p ?
1-1/m
9
PCP Theorem

• NP PCP½log(n),O(1)
• If x?L ?p Prv accepts p 1
• If x?L ?p Prv accepts p ½.
• r log(n) - how many coins does v need for
  checking p
• q O(1) - the number of places v samples p

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What is it good for?

• NP PCP½log(n),O(1)
• ?
• Gap3-SAT?,1?NP-Hard
• PCP is very important for approximate problems

11
History

• AS92 defining PCP
• ALMSS92 the PCP theorem
• Long and complicated proof
• Uses heavily algebraic tools
• Not good parameters
• BGLR93, RS97,AS97
• Better parameters but even more complicated
• D06 simple combinatorial proof
• STOC06 best paper

12
Main Goal

• We will show an overview of the proof
• Gap3-SAT?,1?NP-Hard
• Actually we
• will prove
• hardness of
• a problem,
• Which has an easy reduction to Gap3-SAT

Remember NP PCP½log(n),O(1) ?
Gap3-SATc,1?NP-Hard
13
Constraint Graph (CG)

• Input ltG(V,E), ?, C gt
• G (V,E) undirected graph ,
• ? alphabet,
• C constraints where for every e?E, Ce????.
• Example
• ? 0,1,2

v 0
U 1
U 2
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Constraint Graph (CG)

• Goal
• Find sV-gt? that satisfied all edges
• (u,v)?E s.t. (s(u),s(v))?Cu,v

s(v)1
s(u)1
s(w)1
s(u)0
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CG?NPC

• CG?NP
• Well show reduction
• 3-Col CG
• ? 0,1,2
• For every edge (u,v)?E, Cu,v u ? v

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GapCGa,ß

• Definition
• CG is a constraint graph (like before)
• YES there exists s V-gt? s.t.
• satisfied edges ß.
• NO for any sV-gt?
• satisfied edges ?.

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GapCGa,ß

• Notice
• if G?CG then in every assignment there is
• at least one edge that is not satisfied
• In this case
• GapCG1-1/m,1 CG
• (m E)

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UNSAT(CG)

• UNSAT(CG) the smallest number of
• unsatisfied edges
• all edges
• UNSAT(CG)
• mins(Cu,v (s(u),s(v))?C(u,v)/m)

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UNSAT(CG)

• Remember
• if G?CG
• then in every assignment there is at least
• one edge that is not satisfied
• Therefore
• UNSAT(G)
• m E

1
bad edges


m
all edges
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UNSAT(CG)

• So what do we know?
• It is NP-Hard to distinguish between UNSAT(G)
  0 to UNSAT(G) 1/m
• So what do we need?
• It is NP-Hard to distinguish between UNSAT(G)
  0 to UNSAT(G) ?

21
UNSAT(CG)

• why?
• gt GapCG1- ?,1?NP-Hard
• gt GapCG1- ?,1 p Gap3-SATc,1
• gt Gap3-SATc,1?NP-Hard
• gt PCP theorem!!!

Remember NP PCP½log(n),O(1) ?
Gap3-SATc,1?NP-Hard
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GOAL

• UNSAT(G) 1/m is not enough.
• We want constant ? !!!
• gt UNSAT(G) ?
• How can we increase the percentage of the bad
  edges?

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How?

• Gi ? Gi1
• ?(Gi) ?(Gi1)
• ? UNSAT(Gi1) gt 2 ? UNSAT(Gi)
• Gi1 k? Gi (k is a constant)
• Degree(Gi1) constant

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Road Map

• Preparation
• Degree reduction- Gi ? Gi1 d-regular graph
• Expandering- Gi1 ? Gi2
• Powering- Gi2 ? Gi3
• increase UNSAT(Gi2).
• ? -reduce- Gi3 ? Gi1 back to original ?.

25
Preparation

• After the preparations the graph becomes
• d-regular expander.
• Well see later how to do it.

UNSAT(G) G Deg(G) ?
Constant factor
Constant factor
d-regular
No change
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Powering - overview

• Result UNSAT(Gt) ß ? vt ? UNSAT(G)
• Problem ? becomes ?dt

UNSAT(G) G Deg(G) ?
Constant factor
Constant factor
goes up
goes up (problem!!!)
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Powering

• Let G lt(V,E),?,Cgt a constraint graph.
• We define Gt lt(V,E),?dt,Ctgt
• V V
• E there are k edges between u and v iff there
  are exactly k different t-step paths between them
  in G.

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Powering

• ?dt every u?V has information of its t-step
  neighbors.
• Constraints (u,v)?E will be satisfied if-
• u and v assignment agree about all the shared
  neighbors.
• The assignment satisfies all the shared
  constraints in the original graph G.

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Powering

• Example t 2

Gt
G
?
?
?
?
powering?
?
?
?
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Powering

• t 1 Assignment
• Does it satisfie?

y1 x1 w0
G
Gt
C(1,1)
powering?
C(1,0)
C(0,2)
y1 x1
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Powering

• Lemma (amplification lemma)
• ?ß gt 0,?G lt(V,E),?,Cgt - d-regular-CG
• UNSAT(Gt) ßvt?min(UNSAT(G),1/t)
• UNSAT(Gt) 2?UNSAT(G) or 1/t (constant)

Example If well take t 4/ß2 then
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Powering

• What does it give us?
• UNSAT(G)
• 1/m ? 2/m ? 22/m ?? 2O(log(m))/m ?
• We need to repeat the procedure O(log(m)) times

33
Powering - Summary

• Result UNSAT(Gt) ß ? vt ? UNSAT(G)
• Problem ? becomes ?dt

UNSAT(G) G Deg(G) ?
Constant factor
Constant factor
goes up
goes up (problem!!!)
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Reduce ?

• Problem
• ? is exponential in n - ?dlog(n).
• The reduction wont be polynomial.
• Solution
• There is a codethat reduce ?

UNSAT(G) G Deg(G) ?
no change
by constant
by constant
back to usual
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Summary

• In each iteration (3 steps combined)
• Preparations, Powering and Reduce ?
• Doubles UNSAT
• Increases G by constant factor.
• Increases degree.
• Doesnt change ?.

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More about Degree Reduction

• Lemma 3.2 (constant degree)
• Any G lt(V,E),?,Cgt, CG
• can be transformed to
• d-regular-CG G lt(V,E),?,Cgt
• s.t.
• a?UNSAT(G) UNSAT(G)
• for d gt 0, a gt 0.

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Degree Reduction - Proof

• What is our goal?
• To make the graph d-regular
• a?UNSAT(G) UNSAT(G) for a gt 0

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Degree Reduction - Proof

• For every u?V we add k new vertices ui,
• when u appears k times in the constraints.
• k degree(v)

?
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Degree Reduction - Proof

• What should be the constrains between the ui?
• For any i ? j the constrains are- ui uj
• How will we do it?

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Degree Reduction - Proof

• First try
• Connect them all as a clique!!!
• Whats the problem?
• We have n2 good edges so the
• UNSAT(G) is very small
• Not good for us!!!

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Degree Reduction - Proof

• Second try
• Connect them all in a line
• Whats the problem?
• The UNSAT is still getting smaller!

?
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Degree Reduction - Proof
s(x)1
s(u1)1
C(1,1)

• Example

s(x)1
..
s(y)1
s(ui)1
C(1,1)
C(1,1)
. .
s(u)0
?
C(1,1)
s(z)1
C(0,1)
C(0,1)
s(y)1
..
s(ui1)0
C(0,1)
.
. .
s(w)1
s(z)1
s(un)0
C(0,1)
UNSAT1/3
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Degree Reduction - Proof

• Third and last try
• Connect them all in an Expander!!!
• Reminder

Expander a?SE(S,S) where agt1.
44
Degree Reduction - Proof

• Can we build an Expander?
• Theorem ?n vertices we can build a d-1
• regular expander with a gt 0 and edges
• l?n (linear in n)

45
Degree Reduction - Proof
There is Expander between the Ui cSE(S,S)
S A set S V½

• Why is Expander good?

S
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Degree Reduction - Proof

• If the bad guy
• takes a set
• S of ui
• and cheat
• The number of
• unsatisfied
• edges
• Will grow

S
x
u1
un
u2
u3
u7
Edges outside the Expander
u4
u6
u5
y
z
47
More about Expandering

• Problem
• what if we have this graph?
• F satisfied
• edges
• S unsatisfied
• edges
• C the edges
• between T to S

S
C
F
48
More about Expandering
S

• What will happen
• To Gt (powering)?
• UNSAT(Gt)
• might be smaller
• than UNSAT(G)
• Solution
• expander!! (last week)

C
F
49
More about Expandering

• Definition
• G is expander if for every S in V s.t.
• S V/2, denote E(S,S) as the number of
  edges between S to V/S (S),
• a?S E(S,S), where a gt 1.

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More about Expandering

• Why is expander a solution?
• Denote S as the bad edges (unsatisfied edges).
• In every power iteration S multiplies itself in a
  constant gt 1.
• S?(1a/d)S?(1a/d)2S?
• ?(1a/d)O(log(m))S m
• So after O(log(m)) iterations we get to all the
  edges.

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Preparation - summary

• After the preparations the graph becomes
• d-regular expander.
• Well see later how to do it.

UNSAT(G) G Deg(G) ?
Constant factor
Constant factor
d-regular
No change
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summary

• CG?NPC
• CG GapCG1-1/m,1
• GapCG1- 1/m ,1 ?p(polytime) GapCG?,1
• GapCG?,1?NP-Hard
• GapCG?,1 p Gap3-SAT?,1
• Gap3-SAT?,1?NP-Hard
• PCP theorem!!!