understanding parallel repetition requires understanding foams
DESCRIPTION
Understanding Parallel Repetition Requires Understanding Foams. Uri Feige Microsoft. Guy Kindler Weizmann. Ryan O’Donnell CMU. What we wanted to solve. Strong Parallel Repetition Problem: Let G be a 2-prover 1-round game with answer sets A , B . - PowerPoint PPT PresentationTRANSCRIPT
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Uri FeigeMicrosoft
Understanding Parallel Repetition
Requires Understanding Foams
Guy KindlerWeizmann
Ryan O’DonnellCMU
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What we wanted to solve
Strong Parallel Repetition Problem:
Let be a 2-prover 1-round game with answer sets A, B.
Is it true that val( ) · 1 −
) val( d) · (1 − ())d/log(|A||B|) ?
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A special case
Strong Unique-Games Parallel Repetition Problem:
Let be a 2P1R game with answer sets A, B and unique constraints.
Is it true that val( ) · 1 −
) val( d) · (1 − ())d/log(|A||B|) ?
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A further special case
Strong 2-Lin Parallel Repetition Problem:
Let be a 2P1R game with 2-Lin constraints.
Is it true that val( ) · 1 −
) val( d) · (1 − ())d ?
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A further further special case
Odd-Cycle Parallel Repetition Problem:
Let Cm be the Odd-Cycle game of length m, which satisfies
Is it true that val(Cm) = 1 − (1/m).
Is it true that val(Cmd) · (1 − (1/m))d ?
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Further reduces to
Torus Blocking Problem on (md)1:
Let (md)1 be the “discrete torus graph”:
vertex set = md,
edge set = {(x, y) : ||x − y||1 · 1}.
To block all cycles that “wrap around”, what’s the least fraction of
edges you can delete?
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Our results
• Improved lower bound for Torus Blocking Problem,
which implies
• Improved upper bounds for Odd Cycle Parallel Repetition problem.
• At least, if you look at the parameters in the right way.
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This looks kind of pathetic
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But it’s not our fault
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Further further reduces to
Foam on d / d Problem:
What is the least surface area of a cell which tiles d by d ?
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Further further reduces to
Foam on d / d Problem:
What is the least surface area of a cell which tiles d by d ?
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Kelvin foam
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Similar questions are hard open problems in geometry
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Foam on d / d
Let A(d) denote the least possible surface area…
Upper bound?
A(d ) · d.
Lower bound?
÷ 2.
the unit cube
the volume-1 ball
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Other bounds
• A(d) · d − 2−O(d log d) (put a radius-½ sphere at cube’s corner)
• (the hexagon was optimal [Choe’89])
• For d = 3, nothing known except sphere vs. cube:
2.42 ¼ (9/2)1/3 · A(3) < 3.
Experts’ d = 3 conjecture: same combinatorial structure as “Kelvin Foam”
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A prize
For £100:
Prove or disprove: A(d) ¸ d 1−o(1).
For £25: Prove
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Foams as torus blockers
Take the unit cube in d.
Identify opp. faces so it’s a torus.
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Foams as torus blockers
Take the unit cube in d.
Identify opp. faces so it’s a torus.
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Foams as torus blockers
Take the unit cube in d.
Identify opp. faces so it’s a torus.
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Foams as torus blockers
Take the unit cube in d.
Identify opp. faces so it’s a torus.
To block all cycles that “wrap around”, what’s the least amount
of “wall” (d −1 dimensional surface) you need to build?
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Foams as torus blockers
Take the unit cube in d.
Identify opp. faces so it’s a torus.
To block all cycles that “wrap around”, what’s the least amount
of “wall” (d −1 dimensional surface) you need to build?
(Hence the ÷ 2: surface counted twice – inside and outside.)
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A worse lower bound: ssss
• Wall S at least blocks all axis-parallel cycles.
• So projecting S onto d faces must cover them.
• Let P be a tiny patch on S, with unit normal n.
• Area contributed to projection on ith face:
|h n, eii| area(P)
• Sum over i: Equals (i |ni|) · area(P)
· · area(P) [Cauchy-Schwarz]
• Integrate over P: · · area(S).
• But this contribution better exceed d.P
n
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A worse lower bound: ssss
• Wall S at least blocks all axis-parallel cycles.
• So projecting S onto d faces must cover them.
• Let P be a tiny patch on S, with unit normal n.
• Area contributed to projection on ith face:
|h n, eii| area(P)
• Sum over i: At most hn, (1, …, 1)i area(P)
· · area(P) [Cauchy-Schwarz]
• Integrate over P: · · area(S).
• But this contribution better exceed d. P
n
We already lost here.
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What’s this got to do with Parallel Repetition?
What is Parallel Repetition?
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Bipartite Constraint Graphs
w – a weight
Label Set = { }
– a constraint
not OKOKOKOKnot OKOKOKOKnot OK
The w’s sum up to 1.
Whole thing is called . val( ) denotes max weight simultaneously satisfiable.
X Y1
2
3
4
5
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Bipartite Constraint Graphs
w – a weight
Label Set = { }
– a constraint
not OKOKOKOKnot OKOKOKOKnot OK
The w’s sum up to 1.
Whole thing is called . val( ) denotes max weight simultaneously satisfiable.
X Y1
2
3
4
5
6
7
8
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2-Prover 1-Round Games
in complexity theory
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Bipartite Constraint Graphs
w – a weight
Label Set = { }
– a constraint
not OKOKOKOKnot OKOKOKOKnot OK
The w’s sum up to 1.
Whole thing is called . val( ) denotes max weight simultaneously satisfiable.
X Y1
2
3
4
5
6
7
8
9
10
11
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19
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Nonlocal Games
in foundations of quantum mechanics
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Parallel Repetition: d “rounds”
w – a weight
Label Set = { }
– a constraint
not OKOKOKOKnot OKOKOKOKnot OK
The w’s sum up to 1.
Whole thing is called . val( ) denotes max weight simultaneously satisfiable.
X Y1
2
3
4
5
6
7
8
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Parallel Repetition: d “rounds”
w – a weight
Label Set = { }
– a constraint
not OKOKOKOKnot OKOKOKOKnot OK
The w’s sum up to 1.
Whole thing is called . val( ) denotes max weight simultaneously satisfiable.
Xd Yd
1 8 4 3 8 14 20 13 17 18
d
eg:
weight = w1,14 w8,20 w4,13 w3,17 w8,18
constraint = 1,14 8,20 4,13 3,17 8,18
d
d
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Value under Parallel Repetition
val(d) = val()d ?
val(d) · val() ?
val(2) < val() ?
val(d) ! 0 as d ! 1 ?
false
true
false
true
(took 6 years to prove)
True or False?
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Raz’s Parallel Repetition Theorem
Raz ’95: val( ) · 1 − ) val( d) · (1 − poly()) d/log(# labels)
Tremendously important theorem for proving hardness of approximation results.
Holenstein ’07: poly() can be 3 / 4000.
Strong Parallel Repetition Problem: can this be improved to ()?
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The “2-Lin” special case
# labels = 2, each constraint is either “=” or “”
Feige-Lovász ’91 + Goemans-Williamson ’95:
val( ) · 1 − ) val( d) · (1 − c)) d, where c = 2/4.
Strong 2-Lin Parallel Repetition Problem: Can this be improved to ()?
My conjecture: Yes.
My motivation: Would show that sharp hardness-of-approx for Max-Cut is
“Unique Games Conjecture”-complete,
not just “Unique Games Conjecture”-hard.
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Simplest 2-Lins: The Odd Cycle Games
m nodes ) val = 1 – 1/m
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Simplest 2-Lins: The Odd Cycle Games
1/3 total weight on self-loops ) val = 1 – (2/3)/m
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After Parallel Rep: Discrete Torus Graph
52
NB: Constraints are “unique”
(x,y) an edge iff ||x-y||1 · 1
1st col. diff., 2nd col. same
1st col. diff., 2nd col. diff.
1st col. same, 2nd col. diff.
1st col. same, 2nd col. same
(self-loops, not pictured)
Constraints
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After Parallel Rep: Discrete Torus Graph
52
NB: Constraints are “unique”
(x,y) an edge iff ||x-y||1 · 1
1st col. diff., 2nd col. same
1st col. diff., 2nd col. diff.
1st col. same, 2nd col. diff.
1st col. same, 2nd col. same
(self-loops, not pictured)
Constraints
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After Parallel Rep: Discrete Torus Graph
52
NB: Constraints are “unique”
(x,y) an edge iff ||x-y||1 · 1
Given set of Failure Edges,
there’s a corresp. labeling iff all “topologically nontrivial” cycles blocked (*)
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val(Cmd) vs. Torus Blocking
Basically(*), val(Cmd) = 1 − (d, m),
where (d, m) = least fraction of edges you need to delete from md graph
to eliminate all cycles that “wrap around”.
To prove strong upper bound for val(Cmd),
must prove strong lower bound for (d, m).
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Discrete vs. Continuous Foams
But strong lower bound for (d, m) implies strong lower bound for A(d).
Proposition: Upper bound for A(d) implies upper bound for (d, m).
Specifically, (d, m) · const. A(d) / m.
Proof:
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Discrete vs. Continuous Foams
But strong lower bound for (d, m) implies strong lower bound for A(d).
Proposition: Upper bound for A(d) implies upper bound for (d, m).
Specifically, (d, m) · const. A(d) / m.
Proof:
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Hence the paper’s title
To understand the truth about parallel repetition,
you must get good upper bounds for val(Cmd)
(a special case of a special case of a special case of the general case).
But this requires good lower bounds for the continuous d / d Foam Problem.
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Our results
What do we actually prove in the paper?!
Main Theorem: The continuous foam lower bound
can be discretified into a lower bound for (d, m):
(d, m) ¸ (if d · m2 log m, say).
Hence val(Cmd) · 1 −
Proof: A lot of Fourier analysis.
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Our results
What we got: val(Cmd) · 1 −
Best previously: 1 − (d) ¢ (1/m)2
What we really wanted: 1 − (d) ¢ (1/m)
m = 33
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