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Random non-local games Andris Ambainis, Artūrs Bačkurs, Kaspars Balodis, Dmitry Kravchenko, Juris Smotrovs, Madars Virza University of Latvia

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Non-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 P a, b (x, y) satisfied. Alice, Bob win if a condition P a, b (x, y) satisfied. Alice Bob Referee ab xy

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Example 1 [CHSH] Winning conditions for Alice and Bob Winning conditions for Alice and Bob (a = 0 or b = 0) x = y. (a = 0 or b = 0) x = y. (a = b = 1) x y. (a = b = 1) x y. Alice Bob Referee ab xy

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Example 2 [Cleve et al., 04] Alice and Bob attempt to “prove” that they have a 2-coloring of a 5-cycle; Alice and Bob attempt to “prove” that they have a 2-coloring of a 5-cycle; Referee may ask one question about color of some vertex to each of them. Referee may ask one question about color of some vertex to each of them.

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Example 2 Referee either: asks i th vertex to both Alice and Bob; they win if answers equal. asks i th vertex to both Alice and Bob; they win if answers equal. Asks the i th vertex to Alice, (i+1) st to Bob, they win if answers different. Asks the i th vertex to Alice, (i+1) st to Bob, they win if answers different.

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Non-local games in quantum world Shared quantum state between Alice and Bob: Shared quantum state between Alice and 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 in the classical case.

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Example:CHSH game Winning condition: (a = 0 or b = 0) x = y. (a = 0 or b = 0) x = y. (a = b = 1) x y. (a = b = 1) x y. Winning probability: 0.75 classically. 0.75 classically. 0.85... quantumly. 0.85... quantumly. A simple way to verify quantum mechanics. Alice Bob Referee ab xy

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Example: 2-coloring game Alice and Bob claim to have a 2-coloring of n- cycle, n- odd; Alice and Bob claim to have a 2-coloring of n- cycle, n- odd; 2n pairs of questions by referee. 2n pairs of questions by referee. Winning probability: classically. classically. quantumly. quantumly.

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Random non-local games a, b {1, 2,..., N}; a, b {1, 2,..., N}; x, y {0, 1}; x, y {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 ab xy

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

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The main results Let N be the number of possible questions to Alice and Bob. Let N be the number of possible questions to Alice and Bob. Classical winning probability p cl satisfies Classical winning probability p cl satisfies Quantum winning probability p q satisfies Quantum winning probability p q satisfies

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Another interpretation Value of the game = p win – (1-p win ). Value of the game = p win – (1-p win ). Quantum advantage: Quantum advantage: Corresponds to Bell inequality violation

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Related work Randomized constructions of non-local games/Bell inequalities with large quantum advantage. Junge, Palazuelos, 2010: N questions, N answers, √N/log N advantage. Regev, 2011: √N advantage. Briet, Vidick, 2011: large advantage for XOR games with 3 players.

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Differences JP10, R10, BV11: Goal: maximize quantum-classical gap; Randomized constructions = tool to achieve this goal; This work: Goal: understand the power of quantum strategies in random games;

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Methods: quantum Tsirelson’s theorem, 1980: Alice’s strategy - vectors u 1,..., u N, ||u 1 || =... = ||u N || = 1. Alice’s strategy - vectors u 1,..., u N, ||u 1 || =... = ||u N || = 1. Bob’s strategy - vectors v 1,..., v N, ||v 1 || =... = ||v N || = 1. Bob’s strategy - vectors v 1,..., v N, ||v 1 || =... = ||v N || = 1. Quantum advantage Quantum advantage

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Random matrix question What is the value of What is the value of for a random 1 matrix A? for a random 1 matrix A? Can be upper-bounded using ||A||=(2+o(1)) √N

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Upper bound =

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Upper bound theorem Theorem For a random A, Corollary The advantage achievable by a quantum strategy in a random XOR game is at most

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Lower bound There exists u: There exists u: There are many such u: a subspace of dimension f(n), for any f(n)=o(n). There are many such u: a subspace of dimension f(n), for any f(n)=o(n). Combine them to produce u i, v j : Combine them to produce u i, v j :

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Marčenko-Pastur law Let A – random N·N 1 matrix. W.h.p., all singular values are between 0 and (2 o(1)) √N. Theorem (Marčenko, Pastur, 1967) W.h.p., the fraction of singular values i c √N is

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Modified Marčenko-Pastur law Let e 1, e 2,..., e N be the standard basis. Theorem With probability 1-O(1/N), the projection of e i to the subspace spanned by singular values j c √N is

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Classical results Let N be the number of possible questions to Alice and Bob. Let N be the number of possible questions to Alice and Bob. Theorem Classical winning probability p cl satisfies Theorem Classical winning probability p cl satisfies

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Methods: classical Alice’s strategy - numbers u 1,..., u N {-1, 1}. Alice’s strategy - numbers u 1,..., u N {-1, 1}. Bob’s strategy - numbers Bob’s strategy - numbers v 1,..., v N {-1, 1}. v 1,..., v N {-1, 1}. Classical advantage Classical advantage

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Classical upper bound Chernoff bounds + union bound: Chernoff bounds + union bound: with probability 1-o(1). with probability 1-o(1).

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Classical lower bound Random 1 matrix A; Operations: flip signs in all entries in one column or row; Goal: maximize the sum of entries.

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Greedy strategy Choose u 1,..., u N one by one. 1...1 1 1... k-1 rows that are already chosen 2 0 -2... 0 1 1... 1 1... Choose the option which maximizes agreements of signs

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Analysis On average, the best option agrees with fraction of signs. 1 1... 1 2 0 -2... 0 1... Choose the option which maximizes agreements of signs If the column sum is 0, it always increases.

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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 2 0 -2... 0 1... Choose the option which maximizes agreements of signs Expected distance from origin = 1.27... √N.

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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:

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Comparison Random XOR game: Random XOR game: Biggest gap for XOR games: Biggest gap for XOR games:

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Open problems 1. 1. We have What is the exact order? 2. 2. Gaussian A ij ? Different probability distributions? 3. 3. Random games for other classes of non-local games?

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