Random non-local games
-
Upload
megan-ware -
Category
Documents
-
view
27 -
download
0
description
Transcript of Random non-local games
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: