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Testing Linearity against Non-Signaling Strategies Alessandro Chiesa Peter Manohar Igor Shinkar UC Berkeley What is linearity testing? 1 Linearity Testing 2 Linearity Testing Given oracle access to f:{0,1} n {0,1} decide if: (1) f is

  1. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k 1/3 0 1 1 0 F S 1/2 0 0 1 1 1/6 1 1 0 0 0 1 1 0 S ⊆ {0, 1} n , |S| ≤ k 9

  2. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 0 1 1 0 F S 1/2 0 0 1 1 1/6 1 1 0 0 0 1 1 0 S ⊆ {0, 1} n , |S| ≤ k 9

  3. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 0 1 1 0 S ⊆ {0, 1} n , |S| ≤ k 9

  4. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 S ⋂ T 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 0 1 1 0 S ⊆ {0, 1} n , |S| ≤ k 9

  5. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 S ⋂ T 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 1/3 0 1 1 0 1/2 0 0 1 1 1/6 0 1 1 0 1 1 0 0 S ⊆ {0, 1} n , |S| ≤ k 9

  6. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 S ⋂ T 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 1/3 1 0 1/2 1 1 1/6 0 1 1 0 0 0 S ⊆ {0, 1} n , |S| ≤ k F S | S ⋂ T 9

  7. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 S ⋂ T 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 1/3 1/3 1 0 1 0 0 1 1/2 1/2 1 1 1 1 1 1 1/6 1/6 0 1 1 0 0 0 0 0 0 0 S ⊆ {0, 1} n , |S| ≤ k F S | S ⋂ T 9

  8. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 S ⋂ T 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 1/3 1/3 1 0 1 0 1/2 1/2 1 1 1 1 1/6 1/6 0 1 1 0 0 0 0 0 S ⊆ {0, 1} n , |S| ≤ k F S | S ⋂ T F T | S ⋂ T 9

  9. Non-Signaling Functions Definition: A k-non-signaling function F: {0,1} n → {0,1} is a collection of distributions {F S } S over functions f: S → {0,1}, ∀ S ⊆ {0, 1} n , |S| ≤ k that satisfies the non-signaling property: ∀ S, T ⊆ {0, 1} n , |S|, |T| ≤ k F S | S ⋂ T ≣ F T | S ⋂ T (the marginal distributions are equal) 1/3 1/3 0 1 1 0 1 0 0 1 1/3 0 1 1 0 S ⋂ T 1/2 F S 1/2 0 0 1 1 1 1 1 1 1/2 0 0 1 1 1/6 1/6 1 1 0 0 0 0 0 0 1/6 1 1 0 0 F S F T 1/3 1/3 1 0 1 0 1/2 1/2 1 1 1 1 1/6 1/6 0 1 1 0 0 0 0 0 S ⊆ {0, 1} n , |S| ≤ k ≣ F S | S ⋂ T F T | S ⋂ T 9

  10. NS Linearity Testing 10

  11. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. 10

  12. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: 10

  13. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling Verifier 10

  14. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling Verifier x,y ← {0,1} n 10

  15. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling S = { x,y,x+y } Verifier x,y ← {0,1} n 10

  16. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling S = { x,y,x+y } { F(x), F(y), F(x+y) } Verifier x,y ← {0,1} n 10

  17. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling S = { x,y,x+y } { F(x), F(y), F(x+y) } Verifier x,y ← {0,1} n F(x) + F(y) ?= F(x+y) 10

  18. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling S = { x,y,x+y } { F(x), F(y), F(x+y) } Verifier x,y ← {0,1} n F(x) + F(y) ?= F(x+y) Does the test do anything useful? Pr x,y,F [F passes] = 1 → some global conclusion? 10

  19. NS Linearity Testing Given oracle access to F:{0,1} n → {0,1}, k-non-signaling. Run the same test as before: F: {0,1} n → {0,1} k-non-signaling S = { x,y,x+y } { F(x), F(y), F(x+y) } Verifier x,y ← {0,1} n F(x) + F(y) ?= F(x+y) Does the test do anything useful? Pr x,y,F [F passes] = 1 → some global conclusion? How? NS functions are collections of local distributions. 10

  20. Let’s first understand non-signaling functions 11

  21. Examples of NS Functions 12

  22. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 12

  23. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 12

  24. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 • A more interesting example: F: {1,2,3} → {0,1}, k = 2 F S defined as: 12

  25. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 • A more interesting example: F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} 1/2 0 0 1/2 1 1 12

  26. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 • A more interesting example: F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} 1/2 1/2 0 0 0 0 1/2 1/2 1 1 1 1 12

  27. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 • A more interesting example: F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 12

  28. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 • A more interesting example: F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Cannot be explained by a distribution… 12

  29. Examples of NS Functions • A function: 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 • A distribution over functions: 1/2 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 1/6 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 1/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 • A more interesting example: F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Cannot be explained by a distribution… But can try anyways! 12

  30. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 13

  31. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. 13

  32. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: 13

  33. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} 13

  34. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 13

  35. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 1/2 = Pr[F {1,2} = (0,0)] = q (0,0,0) + q (0,0,1) 13

  36. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 1/2 = Pr[F {1,2} = (0,0)] = q (0,0,0) + q (0,0,1) 1/2 = Pr[F {1,2} = (1,1)] = q (1,1,0) + q (1,1,1) 13

  37. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 1/2 = Pr[F {1,2} = (0,0)] = q (0,0,0) + q (0,0,1) 1/2 = Pr[F {1,2} = (1,1)] = q (1,1,0) + q (1,1,1) 0 = Pr[F {1,2} = (0,1)] = q (0,1,0) + q (1,0,1) 13

  38. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 1/2 = Pr[F {1,2} = (0,0)] = q (0,0,0) + q (0,0,1) 1/2 = Pr[F {1,2} = (1,1)] = q (1,1,0) + q (1,1,1) 0 = Pr[F {1,2} = (0,1)] = q (0,1,0) + q (1,0,1) 0 = Pr[F {1,2} = (1,0)] = q (1,0,0) + q (1,0,1) … 13

  39. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 1/2 = Pr[F {1,2} = (0,0)] = q (0,0,0) + q (0,0,1) 1/2 = Pr[F {1,2} = (1,1)] = q (1,1,0) + q (1,1,1) 0 = Pr[F {1,2} = (0,1)] = q (0,1,0) + q (1,0,1) 0 = Pr[F {1,2} = (1,0)] = q (1,0,0) + q (1,0,1) … Fact: system of linear equations has a solution. 13

  40. Example cont. F: {1,2,3} → {0,1}, k = 2 F S defined as: F {1,2} F {2,3} F {1,3} 1/2 1/2 1/2 0 0 0 0 0 1 1/2 1/2 1/2 1 1 1 1 1 0 Let’s try to write it as a distribution anyways. System of linear eqs: Variables: q f for every f: {1,2,3} → {0,1} Constraints: 1/2 = Pr[F {1,2} = (0,0)] = q (0,0,0) + q (0,0,1) 1/2 = Pr[F {1,2} = (1,1)] = q (1,1,0) + q (1,1,1) 0 = Pr[F {1,2} = (0,1)] = q (0,1,0) + q (1,0,1) 0 = Pr[F {1,2} = (1,0)] = q (1,0,0) + q (1,0,1) … Fact: system of linear equations has a solution. Solution has negative entries, but marginals on “queryable sets” are non-negative. 13

  41. Quasi-Distributions 14

  42. Quasi-Distributions A quasi-distribution is a distribution, only “probabilities” are allowed to be negative. 14

  43. Quasi-Distributions A quasi-distribution is a distribution, only “probabilities” are allowed to be negative. -1/3 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 2/3 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 2/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 14

  44. Quasi-Distributions A quasi-distribution is a distribution, only “probabilities” are allowed to be negative. -1/3 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 2/3 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 2/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 A quasi-distribution is k-local if every marginal on k points is a (standard) distribution. 14

  45. Quasi-Distributions A quasi-distribution is a distribution, only “probabilities” are allowed to be negative. -1/3 1 0 1 1 0 0 0 1 1 1 0 0 1 0 1 1 2/3 1 1 0 0 1 1 0 1 1 0 0 1 1 1 0 1 2/3 0 0 1 1 1 1 1 0 0 1 0 1 1 1 1 0 A quasi-distribution is k-local if every marginal on k points is a (standard) distribution. Observation: k-local k-non-signaling quasi-distributions functions 14

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