Random non-local gamesRandom non-local games

Andris Ambainis, Artūrs Bačkurs, Andris Ambainis, Artūrs Bačkurs, Kaspars Balodis, Dmitry Kravchenko, Kaspars Balodis, Dmitry Kravchenko,

Juris Smotrovs, Madars VirzaJuris Smotrovs, Madars Virza

University of LatviaUniversity of Latvia

Non-local gamesNon-local games

Referee asks questions a, b to Alice, Bob;Referee asks questions a, b to Alice, Bob;Alice and Bob reply by sending x, y;Alice and Bob reply by sending x, y;Alice, Bob win if a condition PAlice, Bob win if a condition Pa, ba, b(x, y) (x, y)

satisfied.satisfied.

Alice Bob

Referee

a b

x y

Example 1Example 1

Winning conditions for Alice and BobWinning conditions for Alice and Bob (a = 0 or b = 0) (a = 0 or b = 0) x = y. x = y. (a = b = 1) (a = b = 1) x x y. y.

Alice Bob

Referee

a b

x y

Example 2Example 2Alice and Bob attempt to Alice and Bob attempt to

“prove” that they have a “prove” that they have a 2-coloring of a 5-cycle;2-coloring of a 5-cycle;

Referee may ask one Referee may ask one question about color of question about color of some vertex to each of some vertex to each of them.them.

Example 2Example 2Referee either:Referee either: asks iasks ith th vertex to both vertex to both

Alice and Bob; they win Alice and Bob; they win if answers equal.if answers equal.

asks the iasks the ith th vertex to vertex to Alice, (i+1)Alice, (i+1)stst to Bob, they to Bob, they win if answers different.win if answers different.

Example 3Example 3

3-SAT formula F = F3-SAT formula F = F11 F F22 ... ... F Fmm..Alice and Bob attempt to prove that they Alice and Bob attempt to prove that they

have xhave x11, ..., x, ..., xnn: F(x: F(x11, ..., x, ..., xnn)=1.)=1.

Example 3Example 3

Referee chooses FReferee chooses Fii and variable x and variable xj j F Fii..

Alice and Bob win if FAlice and Bob win if Fii satisfied and x satisfied and xjj

consistent.consistent.

F = FF = F11 F F22 ... ... F Fmm..

Alice Bob

Referee

i j

all xj, jFixj

Non-local games in quantum worldNon-local games in quantum world

Shared quantum state between Alice and Shared quantum state between Alice and Bob:Bob:Does not allow them to communicate;Does not allow them to communicate;Allows to generate correlated random bits.Allows to generate correlated random bits.

Alice Bob

Corresponds to shared random bits Corresponds to shared random bits

in the classical case.in the classical case.

Example:CHSH gameExample:CHSH game

Winning condition:Winning condition: (a = 0 or b = 0) (a = 0 or b = 0) x = y. x = y. (a = b = 1) (a = b = 1) x x y. y.

Winning probability:Winning probability:0.75 classically.0.75 classically.0.85... quantumly.0.85... quantumly.

A simple way to verify quantum mechanics.

Alice Bob

Referee

a b

x y

Example: 2-coloring gameExample: 2-coloring gameAlice and Bob claim to Alice and Bob claim to

have a 2-coloring of n-have a 2-coloring of n-cycle, n- odd;cycle, n- odd;

2n pairs of questions by 2n pairs of questions by referee.referee.

Winning probability:Winning probability: classically.classically.

quantumly.quantumly.

Random non-local gamesRandom non-local games

a, b a, b {1, 2, ..., N}; {1, 2, ..., N};x, y x, y {0, 1}; {0, 1};Condition P(a, b, x, y) – random;Condition P(a, b, x, y) – random;

Computer experiments: quantum winning probability larger than classical.

Alice Bob

Referee

a b

x y

XOR gamesXOR games

For each (a, b), exactly one of x = y and x For each (a, b), exactly one of x = y and x y is winning outcome for Alice and Bob.y is winning outcome for Alice and Bob.

The main resultsThe main results

Let n be the number of possible questions Let n be the number of possible questions to Alice and Bob.to Alice and Bob.

Classical winning probability pClassical winning probability pclcl satisfies satisfies

Quantum winning probability pQuantum winning probability pqq satisfies satisfies

Another interpretationAnother interpretation

Value of the game = pValue of the game = pwinwin – (1-p – (1-pwinwin).).

Quantum advantage:

Comparison Comparison

Random XOR game:Random XOR game:

CHSH game:CHSH game:

Best XOR game:Best XOR game:

Methods: quantumMethods: quantum

Tsirelson’s theorem, 1980:Tsirelson’s theorem, 1980:Alice’s strategy - vectors uAlice’s strategy - vectors u11, ..., u, ..., uNN, ||, ||

uu11|| = ... = ||u|| = ... = ||uNN|| = 1.|| = 1.

Bob’s strategy - vectors vBob’s strategy - vectors v11, ..., v, ..., vNN, ||, ||

vv11|| = ... = ||v|| = ... = ||vNN|| = 1.|| = 1. Quantum advantageQuantum advantage

Random matrix questionRandom matrix question

What is the value of What is the value of

for a random for a random 1 matrix A?1 matrix A?

Can be upper-bounded by ||A||=(2+o(1)) N √N

Lower boundLower bound

There exists u:There exists u:

There are many such u: a subspace There are many such u: a subspace of dimension f(N), for any f(N)=o(N).of dimension f(N), for any f(N)=o(N).

Combine them to produce uCombine them to produce uii, v, vjj::

Classical resultsClassical results

Let n be the number of possible questions Let n be the number of possible questions to Alice and Bob.to Alice and Bob.

TheoremTheorem Classical winning probability p Classical winning probability pclcl

satisfiessatisfies

Methods: classicalMethods: classical

Alice’s strategy - numbers Alice’s strategy - numbers uu11, ..., u, ..., uNN {-1, 1}. {-1, 1}.

Bob’s strategy - numbers Bob’s strategy - numbers

vv11, ..., v, ..., vNN {-1, 1}. {-1, 1}.Classical advantageClassical advantage

Classical upper boundClassical upper bound

If AIf Aijij – random, A – random, Aijijuuiivvjj – also random. – also random. Sum of independent random variables;Sum of independent random variables; Sum exceeds 1.65... N √N for any uSum exceeds 1.65... N √N for any u ii, v, vjj, with , with

probability o(1/4probability o(1/4nn).). 44NN choices of u choices of uii, v, vjj.. Sum exceeds 1.65... N √N for at least one of Sum exceeds 1.65... N √N for at least one of

them with probability o(1).them with probability o(1).

Classical lower boundClassical lower bound

Given A, change signs of some of its rows so Given A, change signs of some of its rows so that the sum of discrepanciesthat the sum of discrepancies is is maximized, maximized,

Greedy strategy

Choose u1, ..., uN one by one.

1 -1 -1 ... 1

... ... ... ... ...

1 1 -1 ... -1

k-1 rows that are already chosen

2 0 -2 ... 0

-1 1 1 ... 1

1 -1 -1 ... -1

Choose the optionwhich maximizesagreements of signs

Analysis

On average, the best option agrees with fraction of signs.

-1 1 1 ... 1

2 0 -2 ... 0

1 -1 -1 ... -1

Choose the optionwhich maximizesagreements of signs

If the column sum is 0, it always increases.

Rigorous proof

Consider each column separately. Sum of values performs a biased random walk, moving away from 0 with probability in each step.

-1 1 1 ... 1

2 0 -2 ... 0

1 -1 -1 ... -1

Choose the optionwhich maximizesagreements of signs

Expected distance from origin = 1.23... √N.

Conclusion

We studied random XOR games with n questions to Alice and Bob.

For both quantum and classical strategies, the best winning probability ½.

Quantumly:

Classically: