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Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close...

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Shengyu Zhang The Chinese University of Hong Kong
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Page 1: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Shengyu Zhang

The Chinese University of Hong Kong

Page 2: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Why we are here?

Understanding the power of quantum

Computation: quantum algorithms/complexity

Communication: quantum info. theory

This work: game theory

Page 3: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Game: Two basic forms

strategic (normal) form extensive form

Page 4: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Game: Two basic forms

strategic (normal) form

n players: P1, …, Pn

Pi has a set Si of

strategies

Pi has a utility

function ui: S→ℝ

S = S1 S2 ⋯ Sn

Page 5: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Nash equilibrium

Nash equilibrium: each player has adopted

an optimal strategy, provided that others

keep their strategies unchanged

Page 6: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Nash equilibrium

Pure Nash equilibrium:

a joint strategy s = (s1, …, sn) s.t. i,

ui(si,s-i) ≥ ui(si’,s-i)

(Mixed) Nash equilibrium (NE):

a product distribution p = p1 … pn s.t. i,si’

Es←p[ui(si,s-i)] ≥ Es←p[ui(si’,s-i)]

Page 7: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Correlated equilibrium

Correlated equilibrium (CE): p s.t. i, si, si’

CE = NE ∩ product distributions

Es

¡ ià p(¢js

i)[u

i(s

i;s

¡ i)] ¸ E

s¡ i

à p(¢jsi)[u

i(s0

i;s

¡ i)]

Nash and Aumann: two Laureate of Nobel Prize in Economic Sciences

Page 8: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Why correlated equilibrium?

Cross Stop

Cross -100

-100

0

1

Stop 1

0

0

0

Game theory

natural

2 pure NE: one crosses and one stops. Payoff: (0,1) or (1,0) Bad: unfair.

1 mixed NE: both cross w.p. 1/101. Good: Fair

Bad: Low payoff: both ≃ 0.0001

Worse: Positive chance of crash

CE: (Cross,Stop) w.p. ½, (Stop,Cross) w.p. ½ Fair, high payoff, 0 chance of crash.

Traffic Light

Page 9: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Why correlated equilibrium?

Game theory

natural

Math

nice

Set of correlated equilibria is convex.

The NE are vertices of the CE polytope (in any non-

degenerate 2-player game)

All CE in graphical games can be represented by

ones as product functions of each neighborhood.

Page 10: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Why correlated equilibrium?

Game theory

natural

Math

nice

[Obs] A CE can found in poly. time by LP.

natural dynamics → approximate CE.

A CE in graphical games can be found in poly. time.

CS

feasible

Page 11: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

“quantum games”

Non-local games

EWL-quantization of strategic games

J. Eisert, M. Wilkens, M. Lewenstein, Phys. Rev.

Lett., 1999.

Others

Meyer’s Penny Matching

Gutoski-Watrous framework for refereed game

Page 12: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

EWL model

J J-1

Φ1

Φn

⋮⋮ ⋮

s1

sn

|0

|0

What’s this classically?

u1(s)

un(s)

States of concern

Page 13: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

EWL model

Classically we don’t undo the sampling (or

do any re-sampling) after players’ actions.

J J-1

Φ1

Φn

⋮⋮ ⋮

s1

sn

|0

|0

u1(s)

un(s)

Page 14: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

EWL model

J J-1

Φ1

Φn

⋮⋮ ⋮

s1

sn

|0

|0

u1(s)

un(s)

and consider the

state p at this point

Page 15: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Our model

Φ1

Φn

⋮ ⋮

s1

sn

⋮ρ

u1(s)

un(s)

and consider the

state p at this point

A simpler model,

corresponding to classical

games more precisely.

CPTP

Page 16: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Other than the model

Main differences than previous work in

quantum strategic games:

We consider general games of growing sizes.

Previous: specific games, usually 2*2 or 3*3

We study quantitative questions.

Previous work: advantages exist?

Ours: How much can it be?

Page 17: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Central question: How much “advantage” can

playing quantum provide?

Measure 1: Increase of payoff

Measure 2: Hardness of generation

Page 18: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

First measure: increase of payoff

We will define natural correspondences

between classical distributions and quantum

states.

And examine how well the equilibrium

property is preserved.

Page 19: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Quantum equilibrium

classical quantum

Φ1

Φn

⋮ ⋮

s1

sn

⋮ρ

u1(s)

un(s)

C1

Cn

s1’

sn’

⋮p → s

u1(s’)

un(s’)

classical equilibrium:

No player wants to do anything to

the assigned strategy si, if others

do nothing on their parts

- p = p1…pn: Nash equilibrium

- general p: correlated equilibrium

quantum equilibrium:

No player wants to do anything to the

assigned strategy ρ|Hi, if others do nothing

on their parts

- ρ = ρ1… ρn: quantum Nash equilibrium

- general ρ: quantum correlated equilibrium

Page 20: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Correspondence of classical and quantum

states

classical quantum

p: p(s) = ρss

(measure in comp. basis)

ρ

p: distri. on S

ρp = ∑s p(s) |ss| (classical mixture)

|ψp = ∑s√p(s) |s (quantum superposition)

ρ s.t. p(s) = ρss (general class)

Φ1

Φn

⋮ ⋮s1

sn

⋮ρC1

Cn

⋮s1’

sn’⋮p→s

Page 21: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Preservation of equilibrium?

classical quantum

p: p(s) = ρss ρ

p

ρp = ∑s p(s) |ss|

|ψp = ∑s√p(s) |s

ρ s.t. p(s) = ρss

Obs:

ρ is a quantum Nash/correlated equilibrium

p is a (classical) Nash/correlated equilibrium

p NE CE

ρp

|ψp

gen. ρ

Question: Maximum additive and multiplicative increase of payoff

(in a [0,1]-normalized game)?

Page 22: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Maximum additive increase

classical quantum

p: p(s) = ρss ρ

p

ρp = ∑s p(s) |ss|

|ψp = ∑s√p(s) |s

ρ s.t. p(s) = ρss

p NE CE

ρp 0 0

|ψp 0 1-Õ(1/log n)

gen. ρ 1-1/n 1-1/n

Additive

Open: Improve on |ψp?

Question: Maximum additive and multiplicative increase of payoff

(in a [0,1]-normalized game)?

Page 23: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Maximum multiplicative increase

classical quantum

p: p(s) = ρss ρ

p

ρp = ∑s p(s) |ss|

|ψp = ∑s√p(s) |s

ρ s.t. p(s) = ρss

p NE CE

ρp 1 1

|ψp 1 Ω(n0.585…)

gen. ρ n n

multiplicative

Question: Maximum additive and multiplicative increase of payoff

(in a [0,1]-normalized game)?

Open: Improve on |ψp?

Page 24: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Optimization

The maximum increase of payoff on |ψp for a CE p: √pj is short for the column vector (√p1j, …,√pnj)

T.

D ual(A; P ) : min T r (Y ) ¡X

i ; j 2 [n ]

ai j

pi j

(Var : Y 2 Rn £ n )

s.t . Y ºX

j 2 [n ]

ai j

pp

j

pp

jT ; 8i 2 [n]

Non-concavePr imal: max

X

i ;j 2 [n ]

ai j

(p

pj

T Ei

pp

j¡ p

i j) (Var : A; P; E

i2 Rn£ n ; i 2 [n])

s.t . 0 · ai j

· 1; 8i ; j 2 [n] (The game is [0,1]-normalized.)X

i j

pi j

= 1; pi j

¸ 0; 8i ; j 2 [n] (p is a dist ribut ion.)

X

j

ai j

pi j

¸X

j

ai 0j

pi j

; 8i ; i0; j 2 [n] (p is a correlated equilibrium.)

X

i

Ei

= In; E

iº 0; 8i 2 [n] (f E

ig is a POVM measurement .)

Page 25: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Small n and general case

n=2:

Additive: (1/√2) – 1/2 = 0.2071…

Multiplicative: 4/3.

n=3:

Additive: 8/9 – 1/2 = 7/18 = 0.3888...

Multiplicative: 16/9.

General n:

Tensor product

Carefully designed base case

Page 26: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Second measure: hardness of generation

Why care about generation?

Recall the good properties of CE.

But someone has to generate the correlation.

Also very interesting on its own Bell’s inequality

Game theory

natural

Math

nice

CS

feasible

Page 27: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Correlation complexity

Two players want to share a correlation.

Need: shared resource or communication.

Nonlocality? Comm. Comp.? No private inputs here!

Corr(p) = min shared resource needed QCorr(p): entanglement RCorr(p): public coins

Comm(p) = min communication needed QComm(p): qubits RComm(p): bits

Alice Bob

s rB

x y

r

(x,y) p

rA t

Page 28: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Correlation complexity back in games

Φ1

Φn

⋮ ⋮

s1

sn

⋮seed

u1(s)

un(s)

Correlated

Equilibrium

Page 29: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Correlation complexity

Question: Does quantum entanglement have

advantage over classical randomness in

generating correlation?

[Obs] Comm(p) ≤ Corr(p) ≤ size(p)

Right inequality: share the target correlation.

So unlike non-local games, one can always

simulate the quantum correlation by classical.

The question is the efficiency.

Alice Bob

(x,y) p

r

x y

rBrA

size(p) = length of string (x,y)

complexity-version of Bell’s Theorem

Page 30: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Separation

[Thm] p=(X,Y) of size n, s.t.

QCorr(p) = 1, RComm(p) ≥ log(n).

n

[Conj] A random with

Page 31: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Tools: rank and nonnegative rank

[Thm]

P = [p(x,y)]x,y

[Thm]

rank(M) = min r: M = ∑k=1…r Mk, rank(Mk)=1

Nonnegative rank (of a nonnegative matrix): rank+(M) = min r: M = ∑k=1…r Mk, rank(Mk)=1, Mk ≥ 0

Extensively-studied in linear algebra and

engineering. Many connections to (T)CS.

RComm(p) = RCorr(p) = dlog2

rank+

(P)e

1

4log

2rank(P) · QCorr(p) · min

Q: Q± ¹Q= P

log2

rank(Q)

entrywise

Page 32: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Explicit instances

Euclidean Distance Matrix (EDM):

Q(i,j) = ci – cj

where c1, …, cNℝ.

rank(Q) = 2.

[Thm, BL09] rank+(Q∘Q) ≥ log2N

[Conj, BL09] rank+(Q∘Q) = N.

(Even existing one Q implies 1 vs. n separation,

the strongest possible)

Page 33: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Conclusion

Model: natural, simple, rich Non-convex programming; rank+; comm. comp.

Next directions: Improve the bounds (in both measures)

Efficient testing of QNE/QCE?

QCE ← natural quantum dynamics?

Approximate Correlation complexity

[Shi-Z] p: QCorrε(p) = O(log n), RCommε(p) = Ω(√n)

Characterize QCorr?

Mutual info? No! p: Ip = O(n-1/4), QCorr(p) = Θ(log n)

Page 34: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

General n

Construction: Tensor product.

[Lem]

game (u1,u2) (u1’,u2’) (u1u1’,u2u2’)

CE p p’ p p’

old u. u1(|ψp) = u u1(|ψp’) = u’ u∙u’

new

utility

u1(Φ|ψp)

= unew

u1(Φ’|ψp’)

= u’new

u1((Φ⊗Φ’)(|ψp⊗|ψp’))

= unewu’new

Page 35: Shengyu Zhang · 2011-01-10 · 2 log2(n)-ε 1 log2(n) = 1/poly(n). Need: ε 2 and ε 1 very close to 1, yet still admitting a gap of ≈1 when taking power. New construction: Worse

Base case: additive increase

Using the result of n=2?

Additive: ε2log2(n)-ε1

log2(n) = 1/poly(n).

Need: ε2 and ε1 very close to 1, yet still

admitting a gap of ≈ 1 when taking power.

New construction:

Worse than constant

P =

·sin2(²) cos2(²) sin2(²)

0 cos4(²)

¸

U1

=

·cos(²) ¡ sin(²)

sin(²) cos(²)

¸

New u = (1 ¡ sin4(²))log2

n

= 1 ¡ ~O(1=logn)

Old u = (1 ¡ sin2(2²)=4)log2

n

= ~O(1=logn)


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