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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 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 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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Example 3 3-SAT formula F = F 1 F 2 ... F m. 3-SAT formula F = F 1 F 2 ... F m. Alice and Bob attempt to prove that they have x 1,..., x n : F(x 1,..., x n )=1. Alice and Bob attempt to prove that they have x 1,..., x n : F(x 1,..., x n )=1.

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Example 3 Referee chooses F i and variable x j F i. Referee chooses F i and variable x j F i. Alice and Bob win if F i satisfied and x j consistent. Alice and Bob win if F i satisfied and x j consistent. F = F 1 F 2 ... F m. Alice Bob Referee ij all x j, j F i xjxj

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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 classically quantumly 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:

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Comparison Random XOR game: Random XOR game: CHSH game: CHSH game: Best XOR game: Best XOR game:

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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 by ||A||=(2+o(1)) N √N

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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 :

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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 If A ij – random, A ij u i v j – also random. If A ij – random, A ij u i v j – also random. Sum of independent random variables; Sum of independent random variables; Sum exceeds N √N for any u i, v j, with probability o(1/4 n ). Sum exceeds N √N for any u i, v j, with probability o(1/4 n ). 4 N choices of u i, v j. 4 N choices of u i, v j. Sum exceeds N √N for at least one of them with probability o(1). Sum exceeds N √N for at least one of them with probability o(1).

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Classical lower bound Given A, change signs of some of its rows so that the sum of discrepanciesis maximized, Given A, change signs of some of its rows so that the sum of discrepanciesis maximized,

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

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Analysis On average, the best option agrees with fraction of signs 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 Choose the option which maximizes agreements of signs Expected distance from origin = √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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