Equality Alone Does not Simulate Randomness Marc Vinyals Tata - - PowerPoint PPT Presentation
Equality Alone Does not Simulate Randomness Marc Vinyals Tata Institute of Fundamental Research Mumbai, India Joint work with Arkadev Chattopadhyay and Shachar Lovett 34th Computational Complexity Conference Overview Proof Sketch Hierarchy
Marc Vinyals
Tata Institute of Fundamental Research Mumbai, India
Joint work with Arkadev Chattopadhyay and Shachar Lovett 34th Computational Complexity Conference
Overview Proof Sketch Hierarchy
Alice Bob
x y m1 m2 m3 m4 f(x,y)?
Marc Vinyals (TIFR) Equality Alone Does not Simulate Randomness 1 / 14
Overview Proof Sketch Hierarchy
Alice Bob
x y 6 | x + y?
Marc Vinyals (TIFR) Equality Alone Does not Simulate Randomness 1 / 14
Overview Proof Sketch Hierarchy
Alice Bob
x y x (mod 2) 6 | x + y?
Marc Vinyals (TIFR) Equality Alone Does not Simulate Randomness 1 / 14
Overview Proof Sketch Hierarchy
Alice Bob
x y x (mod 2)
No / Maybe, y (mod 3)
6 | x + y?
Marc Vinyals (TIFR) Equality Alone Does not Simulate Randomness 1 / 14
Overview Proof Sketch Hierarchy
Alice Bob
x y x (mod 2)
No / Maybe, y (mod 3) Yes / No
6 | x + y?
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Overview Proof Sketch Hierarchy
Alice Bob
x y x = y?
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Overview Proof Sketch Hierarchy
Alice Bob
x y
x Yes / No
x = y? ◮ Equality needs n + 1 bits.
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r Prr[error] < 1/3
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r
Sample p among first Θ(n) primes
x (mod p)
Yes / No
x = y?
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r ◮ Can solve equality with O(logn) bits.
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r ◮ Can solve equality with O(1) bits.
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r ◮ Can solve equality with O(1) bits. ◮ Greater-than
◮ x ≥ y? ◮ O(logn) bits.
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r ◮ Can solve equality with O(1) bits. ◮ Greater-than
◮ x ≥ y? ◮ O(logn) bits.
◮ Small-set disjointness
◮ x ∩ y = ?, promise |x|,|y| ≤ k ◮ O(k) bits.
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Overview Proof Sketch Hierarchy
Alice Bob
x,r y,r ◮ Can solve equality with O(1) bits. ◮ Greater-than
◮ x ≥ y? ◮ O(logn) bits.
◮ Small-set disjointness
◮ x ∩ y = ?, promise |x|,|y| ≤ k ◮ O(k) bits.
◮ Hashing / Equality is enough to efficiently solve all of these.
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Overview Proof Sketch Hierarchy
[Babai, Frankl, Simon ’86]
Alice Bob Oracle
x, r y, r ◮ Send f(x), g(y) to oracle ◮ Both parties see answer ◮ Cost number of calls
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Overview Proof Sketch Hierarchy
[Babai, Frankl, Simon ’86]
Alice Bob Oracle
x, r y, r
π2/6 ◮ Send f(x), g(y) to oracle ◮ Both parties see answer ◮ Cost number of calls
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Overview Proof Sketch Hierarchy
[Babai, Frankl, Simon ’86]
Alice Bob Oracle
x, r y, r
π2/6 1 1 ◮ Send f(x), g(y) to oracle ◮ Both parties see answer ◮ Cost number of calls
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Overview Proof Sketch Hierarchy
[Babai, Frankl, Simon ’86]
Alice Bob Oracle
x, r y, r
π2/6 1 1 eπ πe ◮ Send f(x), g(y) to oracle ◮ Both parties see answer ◮ Cost number of calls
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Overview Proof Sketch Hierarchy
Question
For every function, is PEQ cost ≃ BPP cost?
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Overview Proof Sketch Hierarchy
Question
For every function, is PEQ cost ≃ BPP cost?
◮ Known false for partial functions ◮ e.g. Maj(x ⊕ y), promise x ⊕ y has either 2n/3 0s or 2n/3 1s. ◮ 2-bit BPP protocol
◮ Sample i ∈ [n] ◮ Send xi ◮ Answer xi ⊕ yi
◮ PEQ cost Ω(n) [Papakonstantinou, Scheder, Song ’14].
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Overview Proof Sketch Hierarchy
Question
For every total function, is PEQ cost ≃ BPP cost?
◮ Known false for partial functions ◮ e.g. Maj(x ⊕ y), promise x ⊕ y has either 2n/3 0s or 2n/3 1s.
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Overview Proof Sketch Hierarchy
Question
For every total function, is PEQ cost ≃ BPP cost?
◮ Known false for partial functions ◮ e.g. Maj(x ⊕ y), promise x ⊕ y has either 2n/3 0s or 2n/3 1s.
Our result: No.
Theorem
There is a total function with BPP cost O(logn) and PEQ cost Ω(n).
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Overview Proof Sketch Hierarchy
Parameters t small constant, n growing, N = 2n/t−1 Input t integers in [−N,N] Alice x = x1,...,xt Bob y = y1,...,yt Output IIP(x,y) = 〈x,y〉 = 0 =
if x1y1 + ··· + xtyt = 0
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Overview Proof Sketch Hierarchy
t small constant, n growing, N = 2n/t−1 IIP(x,y) = x1y1 + ··· + xtyt = 0
Protocol
◮ Sample p among first Θ(n) primes ◮ Send x1 (mod p),...,xt (mod p) ◮ Answer 〈x,y〉 ≡ 0 (mod p)
Cost tlogp = O(logn) Correct with probability 3/4
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y ◮ Prove for PGT. ◮ Can simulate EQ with 2 calls to GT.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y ◮ Prove for PGT. ◮ Can simulate EQ with 2 calls to GT. ◮ Cannot use BPP techniques.
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Overview Proof Sketch Hierarchy
Alice Bob
x y 6 | x + y?
6 6 2 2 8 8 4 4 10 10 1 1 7 7 3 3 9 9 5 5 11 11
◮ Each bit splits inputs into 2 rectangles. ◮ After c bits have 2c rectangles.
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Overview Proof Sketch Hierarchy
Alice Bob
x y x (mod 2) 6 | x + y?
6 6 2 2 8 8 4 4 10 10 1 1 7 7 3 3 9 9 5 5 11 11
◮ Each bit splits inputs into 2 rectangles. ◮ After c bits have 2c rectangles.
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Overview Proof Sketch Hierarchy
Alice Bob
x y x (mod 2)
No / Maybe, y (mod 3)
6 | x + y?
6 6 2 2 8 8 4 4 10 10 1 1 7 7 3 3 9 9 5 5 11 11
◮ Each bit splits inputs into 2 rectangles. ◮ After c bits have 2c rectangles.
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Overview Proof Sketch Hierarchy
Alice Bob
x y x (mod 2)
No / Maybe, y (mod 3) Yes / No
6 | x + y?
6 6 2 2 8 8 4 4 10 10 1 1 7 7 3 3 9 9 5 5 11 11
◮ Each bit splits inputs into 2 rectangles. ◮ After c bits have 2c rectangles.
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Overview Proof Sketch Hierarchy
Alice Bob
x y
1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11
◮ Each bit splits inputs into 2 rectangles. ◮ After c bits have 2c rectangles. ◮ Can show EQ requires 2n rectangles.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) f(x) ≥ g(y) f(x) ≥ g(y) f(x) g(y) ◮ Each call splits inputs into 2 triangles.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) f(x) ≥ g(y) f(x) ≥ g(y) f(x) g(y) ◮ Each call splits inputs into 2 triangles.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) f(x) ≥ g(y) f(x) ≥ g(y) f(x) g(y) ◮ Each call splits inputs into 2 triangles. ◮ After c calls have 2c ?? ◮ Intersections of triangles not triangles. ◮ Each call may use different order.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) f(x) ≥ g(y) f(x) ≥ g(y) f(x) g(y) ◮ Refine partition for free.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) (x,y) ∈ R42 (x,y) ∈ R42 f(x) g(y) ◮ Refine partition for free.
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) (x,y) ∈ R42 (x,y) ∈ R42 f(x) g(y) ◮ Refine partition for free. ◮ Each call splits inputs into 2n
rectangles.
◮ After c calls have 2cn rectangles. ◮ Useless?!
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Overview Proof Sketch Hierarchy
Alice Bob Oracle
x y f(x) g(y) (x,y) ∈ R42 (x,y) ∈ R42 f(x) g(y) ◮ Refine partition for free. ◮ Each call splits inputs into 2n
rectangles.
◮ After c calls have 2cn rectangles. ◮ Useless?! ◮ Many of these rectangles are large.
Can we exploit this?
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Overview Proof Sketch Hierarchy
Total perimeter
|A| + |B| 2n + 2n
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Overview Proof Sketch Hierarchy
Total perimeter
|A| + |B| 2n · 2 · 1/2
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Overview Proof Sketch Hierarchy
Total perimeter
|A| + |B| 2n · (2 · 1/2 + 4 · 1/4)
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Overview Proof Sketch Hierarchy
Total perimeter
|A| + |B| 2n(2 · 1/2 + 4 · 1/4 + ··· + 2n · 2−n) = 2n · n
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Overview Proof Sketch Hierarchy
Total perimeter
|A| + |B|
Greater-than
2n · n
Inner product over 2
(2n − 1) · (2n−1 + 1) ≃ 22n−1
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Overview Proof Sketch Hierarchy
Total η-area
(|A||B|)η 1/2 < η < 1 22ηn(1 · (1/4)η + 2 · (1/16)η + ··· + 2n−1 · 2−2ηn) = 22ηn · q
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Overview Proof Sketch Hierarchy
Theorem
The PGT cost of IIP6 is Ω(n).
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Overview Proof Sketch Hierarchy
Theorem
The PGT cost of IIP6 is Ω(n/logn).
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Overview Proof Sketch Hierarchy
Theorem
The PGT cost of IIP6 is Ω(n/logn). Claim Each call increases perimeter by factor n. After c calls total perimeter 2n · nc.
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Overview Proof Sketch Hierarchy
Theorem
The PGT cost of IIP6 is Ω(n/logn). Claim Each call increases perimeter by factor n. After c calls total perimeter 2n · nc.
Lemma IIP6 has perimeter 2n · exp(Ω(n)).
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Overview Proof Sketch Hierarchy
Theorem
The PGT cost of IIP6 is Ω(n/logn). Claim Each call increases perimeter by factor n. After c calls total perimeter 2n · nc.
Lemma IIP6 has perimeter 2n · exp(Ω(n)).
Claim A function with 1-mass α and 1-rectangles of size at most β has perimeter α/
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Overview Proof Sketch Hierarchy
Theorem
The PGT cost of IIP6 is Ω(n/logn). Claim Each call increases perimeter by factor n. After c calls total perimeter 2n · nc.
Lemma IIP6 has perimeter 2n · exp(Ω(n)).
Claim A function with 1-mass α and 1-rectangles of size at most β has perimeter α/
Claim IIP6 has 1-mass at least ≥ 22n/N2. Claim IIP6 has all 1-rectangles of size at most N6.
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Overview Proof Sketch Hierarchy
◮ What if we had an IIP oracle?
Alice Bob Oracle
x y
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Overview Proof Sketch Hierarchy
◮ What if we had an IIP oracle?
Alice Bob Oracle
x y Theorem
For each t exists t′ such that PIIPt cost of IIPt′ is Ω(n) PEQ PIIPt1 PIIPt2 ··· BPP
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Remarks
◮ PEQ = BPP even for total functions ◮ Hierarchy PEQ PIIPt1 PIIPt2 ··· BPP
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Remarks
◮ PEQ = BPP even for total functions ◮ Hierarchy PEQ PIIPt1 PIIPt2 ··· BPP
Open problems
◮ Is BPP ⊂ PNP? (for total functions) ◮ In particular do BPP functions always have large rectangles?
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Remarks
◮ PEQ = BPP even for total functions ◮ Hierarchy PEQ PIIPt1 PIIPt2 ··· BPP
Open problems
◮ Is BPP ⊂ PNP? (for total functions) ◮ In particular do BPP functions always have large rectangles?
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