Uri FeigeMicrosoft

Understanding Parallel Repetition

Requires Understanding Foams

Guy KindlerWeizmann

Ryan O’DonnellCMU

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|) ?

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|) ?

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 ?

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 ?

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?

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.

This looks kind of pathetic

But it’s not our fault

Further further reduces to

Foam on d / d Problem:

What is the least surface area of a cell which tiles d by d ?

Further further reduces to

Foam on d / d Problem:

What is the least surface area of a cell which tiles d by d ?

Kelvin foam

Similar questions are hard open problems in geometry

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

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”

A prize

For £100:

Prove or disprove: A(d) ¸ d 1−o(1).

For £25: Prove

Foams as torus blockers

Take the unit cube in d.

Identify opp. faces so it’s a torus.

Foams as torus blockers

Take the unit cube in d.

Identify opp. faces so it’s a torus.

Foams as torus blockers

Take the unit cube in d.

Identify opp. faces so it’s a torus.

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?

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.)

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

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.

What’s this got to do with Parallel Repetition?

What is Parallel Repetition?

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

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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

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2-Prover 1-Round Games

in complexity theory

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

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Nonlocal Games

in foundations of quantum mechanics

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

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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

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?

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 ()?

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.

Simplest 2-Lins: The Odd Cycle Games

m nodes ) val = 1 – 1/m

Simplest 2-Lins: The Odd Cycle Games

1/3 total weight on self-loops ) val = 1 – (2/3)/m

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

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

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 (*)

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).

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:

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:

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.

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.

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