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Ryan O’Donnell & Yi Wu Carnegie Mellon University (aka, Conditional hardness for satisfiable 3-CSPs)

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Zwick’s Conjecture [1997] NP ⊆ naPCP 1,5/8+ (O(log n), 3). “ Every language in NP has a probabilistically checkable proof system of polynomial size in which the verifier queries 3 bits of the proof nonadaptively, accepts correct proofs with probability 1, and accepts incorrect proofs with probability at most 5/8+. ” For all > 0,

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Zwick’s Conjecture [1997] NP ⊆ naPCP 1,5/8+ (O(log n), 3). “Given a satisfiable 3CSP, it’s NP-hard to satisfy 5/8+ of the constraints.”

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3CSPs ¢ ¢ ¢

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Zwick’s Conjecture [1997] NP ⊆ naPCP 1,5/8+ (O(log n), 3). 3: minimal. 1: natural, for proof systems. na: natural, for CSP inapproximability. 5/8:this is the conjecture.

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Approximating Satisfiable 3CSPs: 01 [Cook71] NP-hard [Johnson73] In BPP 1/8

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Approximating Satisfiable 3CSPs: NP-hard In BPP 1/8.999999 [AS92,ALMSS92][Johnson73]

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Approximating Satisfiable 3CSPs: NP-hard In BPP.299.8999 [BGS95]

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Approximating Satisfiable 3CSPs: NP-hard In BPP.367.8999 [TSSW96][BGS95]

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Approximating Satisfiable 3CSPs: NP-hard In BPP.3673/4+ [Håstad97][TSSW96]

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Approximating Satisfiable 3CSPs: NP-hard In BPP.5143/4+ [Håstad97][Trevisan97]

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Approximating Satisfiable 3CSPs: NP-hard In BPP 5/83/4+ [Håstad97][Zwick97]

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Zwick’s Conjecture [1997] NP ⊆ naPCP 1,5/8+ (O(log n), 3). “Given a satisfiable 3CSP, it’s NP-hard to satisfy 5/8+ of the constraints.”

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Approximating Satisfiable 3CSPs: NP-hard In BPP 5/83/4+ [Håstad97][Zwick97]

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Approximating Satisfiable 3CSPs: NP-hard In BPP 5/820/27+ [KS06][Zwick97]

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Approximating Satisfiable 3CSPs: NP-hard In BPP 5/820/27+ [KS06][Zwick97] [OW09] Assuming any Khot “D-to-1 Conjecture”

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“Remind me of Khot’s D-to-1 Conjectures?”

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Label-Cover: π 1 (V 1 ) = U 4 π 2 (V 3 ) = U 2 π 3 (V 3 ) = U 9 π 4 (V 2 ) = U 4 Input: U i vbls over [m] V i vbls over [Dm] π j maps are D-to-1 Raz’s Theorem: ∀ δ > 0, if m = poly(1/δ) and D = poly(1/δ), then NP-hard to tell sat’ble from δ-sat’ble.

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Label-Cover: π 1 (V 1 ) = U 4 π 2 (V 3 ) = U 2 π 3 (V 3 ) = U 9 π 4 (V 2 ) = U 4 Input: U i vbls over [m] V i vbls over [Dm] π j maps are D-to-1 2-to-1 Conjecture [Khot02]: ∀ δ > 0, if m = poly(1/δ) and D = 2, then NP-hard to tell sat’ble from δ-sat’ble.

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Label-Cover: π 1 (V 1 ) = U 4 π 2 (V 3 ) = U 2 π 3 (V 3 ) = U 9 π 4 (V 2 ) = U 4 Input: U i vbls over [m] V i vbls over [Dm] π j maps are D-to-1 3-to-1 Conjecture [Khot02]: ∀ δ > 0, if m = poly(1/δ) and D = 3, then NP-hard to tell sat’ble from δ-sat’ble.

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Label-Cover: π 1 (V 1 ) = U 4 π 2 (V 3 ) = U 2 π 3 (V 3 ) = U 9 π 4 (V 2 ) = U 4 Input: U i vbls over [m] V i vbls over [Dm] π j maps are D-to-1 100-to-1 Conjecture [Khot02]: ∀ δ > 0, if m = poly(1/δ) and D = 100, then NP-hard to tell sat’ble from δ-sat’ble.

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Label-Cover: π 1 (V 1 ) = U 4 π 2 (V 3 ) = U 2 π 3 (V 3 ) = U 9 π 4 (V 2 ) = U 4 Input: U i vbls over [m] V i vbls over [Dm] π j maps are D-to-1 Unique Games Conjecture [Khot02]: ∀ δ > 0, if m = poly(1/δ) and D = 1, NP-hard to tell (1−δ)-sat’ble from δ-sat’ble.

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Conjectures: 2-to-1 ⇒ 3-to-1 ⇒ 4-to-1 ⇒ ⇒ 100-to-1 ⇒ ⇒ poly(1/δ)-to-1 (true) None known comparable with UGC.

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Why not use [Raghavendra08]? UGC-based, hence can’t address theory of satisfiable instances Shows Alg/UGC-hard match at some number – but what is the number?

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Why not use [Raghavendra08]? Shows Alg/UGC-hard match at some number – but what is the number? UGC-based, hence can’t address theory of satisfiable instances

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Challenges Shows Alg/UGC-hard match at some number – but what is the number? UGC-based, hence can’t address theory of satisfiable instances ⇒ Design a 1 vs. 5/8 Dictator Test for D-to-1 ⇒ Overcome perfect correlation, pairwise dependence in Invariance Principle arguments

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Take-home message If you can design a Dictator Test that seems to work… … you can make it work.

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Challenges Design a 1 vs. 5/8 Dictator Test for D-to-1 Overcome perfect correlation, pairwise dependence in Invariance Principle arguments

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Challenges Design a 1 vs. 5/8 Dictator Test for D-to-1 Overcome perfect correlation, pairwise dependence in Invariance Principle arguments

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Designing a 1 vs. 5/8 Dictator Test Q: Why is Zwick’s 3CSP alg. stuck at 5/8? A: The “Not-Two” predicate: Linear/random: 5/8 SDP alg: 5/8

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Designing a 1 vs. 5/8 Dictator Test Q: (Håstad01) 1 vs. 5/8+ hardness for NTW? The “Not-Two” predicate:

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Designing a 1 vs. 5/8 Dictator Test Q: (Håstad01) 1 vs. 5/8+ hardness for NTW? A: (Our main thm.) Yes, assuming D-to-1 Conj. for any const. D < ∞.

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Designing a 1 vs. 5/8 Dictator Test [Hås97] gave a 1− vs. 1/2+ Dictator Test for D-to-1, D arbitrary, using XOR. We give a 1 vs. 5/8+ Dictator Test for D-to-1, D constant, using NTW.

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D-to-1 Dictator Tests using Φ Somehow pick corr’d strings: x ∈ {0,1} m, y,z ∈ {0,1} Dm. Test whether Φ( f(x), g(y), g(z) ). x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110100 f ( g (g ( g (g ( m D = 3

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D-to-1 Dictator Tests using Φ Completeness: If f is i th Dictator, g is j th Dictator, j matches i, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≥ c. Soundness: If f and g odd, T 1−η f and T 1−η g have no matching influential variables in common, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≤ s+.

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Completeness: If f is i th Dictator, g is j th Dictator, j matches i, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≥ c. Soundness: If f and g odd, T 1−η f and T 1−η g have no matching influential variables in common, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≤ s+. Håstad: Φ = XOR, c = 1−, s = 1/2

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Completeness: If f is i th Dictator, g is j th Dictator, j matches i, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≥ c. Soundness: If f and g odd, T 1−η f and T 1−η g have no matching influential variables in common, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≤ s+. Us: Φ = NTW, c = 1, s = 5/8

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Håstad’s Dictator Test using XOR Somehow pick corr’d strings: x ∈ {0,1} m, y,z ∈ {0,1} Dm. Test whether XOR( f(x), g(y), g(z) ). x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110100 f ( g (g ( g (g ( m D = 3

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Håstad’s Dictator Test using XOR Pick blocks from some dist. on {0,1} x {0,1} D x {0,1} D. Test whether XOR( f(x), g(y), g(z) ). x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110100 f ( g (g ( g (g ( m D = 3

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Håstad’s Dictator Test using XOR Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Test whether XOR( f(x), g(y), g(z) ). x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110100 f ( g (g ( g (g ( m D = 3

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Håstad’s Dictator Test using XOR Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: Rerandomize each z i with prob. 2. x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110101 f ( g (g ( g (g ( m D = 3

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Completeness: If f is i th Dictator, g is j th Dictator, j matches i, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≥ c. Soundness: If f and g odd, T 1−η f and T 1−η g have no matching influential variables in common, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≤ s+. Håstad: Φ = XOR, c = 1−, s = 1/2

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Completeness: Each “column” (x i,y j,z j ) satisfies XOR w.p. 1−. Soundness: If f and g odd, T 1−η f and T 1−η g have no matching influential variables in common, then Pr x,y,z [ Φ( f(x), g(y), g(z) ) ] ≤ s+. Håstad: Φ = XOR, c = 1−, s = 1/2

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Completeness: Each “column” (x i,y j,z j ) satisfies XOR w.p. 1−. Soundness: “Seems like it should work.” If f = g = Majority, or f = g = Parity then Pr x,y,z [ XOR( f(x), g(y), g(z) ) ] ≤ 1/2+ o(1). Håstad: Φ = XOR, c = 1−, s = 1/2

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Håstad’s Dictator Test using XOR At a technical level… Håstad does direct Fourier Analysis. We sketch an Invariance Principle proof which works for D = O(1). (Have to reprove [MOO05,Mos08].)

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Håstad’s Dictator Test using XOR Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: Rerandomize each z i with prob. 2. Within a block, imperfect correlation between x & (y,z). So E[ f(x) g(y) g(z) ] ≈ E[ Tf(x) Tg(y) Tg(z) ]. x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110101 f ( g (g ( g (g (

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Håstad’s Dictator Test using XOR Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: Rerandomize each z i with prob. 2. Invariance: Tf, Tg have no matching influential vbl. ⇒ can change dist. to anything w/ same 2-wise corr’s. x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110101 f ( g (g ( g (g (

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Håstad’s Dictator Test using XOR Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Then rerandomize x 1. Tweak: Rerandomize each z i with prob. 2. E[ Tf(x) Tg(y) Tg(z) ] = E[ Tf(x) ] E[ Tg(y) Tg(z) ] = 0. x=x= y=y= z=z= 1 101 100011000010 1 001 100000010110101 f ( g (g ( g (g (

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Completeness: Each “column” (x i,y j,z j ) satisfies XOR w.p. 1−. Soundness: “Seems like it should work.” If f = g = Majority, or f = g = Parity then E[ f(x) g(y) g(z) ] ≈ 0. Håstad: Φ = XOR, c = 1−, s = 1/2

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Completeness: Each “column” (x i,y j,z j ) satisfies NTW w.p. 1. Soundness: “Seems like it should work.” If f = g = Majority, or f = g = Parity then E[ f(x) g(y) g(z) ] ≈ 0. Us: Φ = NTW, c = 1, s = 5/8

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Håstad’s Dictator Test using XOR Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. For NTW completeness, okay if “column” is (0,0,0). x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. x=x= y=y= z=z= 1 011 100011000010 0 001 100000010110100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. Unfortunately: perfect correlation between x & (y,z). (Even for D = 2. Imperfect for D = 1, hence [OW09a].) x=x= y=y= z=z= 1 011 101011000010 0 001 101000010100100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. Escape hatch: imperfect correlation between (x,y) & z. So E[ f(x) g(y) g(z) ] ≈ E[ f(x) g(y) Tg(z) ]. x=x= y=y= z=z= 1 011 101011000010 0 001 101000010100100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. Escape hatch: imperfect correlation between (x,y) & z. So E[ f(x) g(y) g(z) ] ≈ E[ f(x) Tg(y) Tg(z) ]. x=x= y=y= z=z= 1 011 101011000010 0 001 101000010100100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. So E[ f(x) g(y) g(z) ] ≈ E[ f(x) Tg(y) Tg(z) ]. Now noise on (y,z) ⇒ as if imperfect corr. for x & (y,z). x=x= y=y= z=z= 1 011 101011000010 0 001 101000010100100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. So E[ f(x) g(y) g(z) ] ≈ E[ f(x) Tg(y) Tg(z) ]. ≈ E[ Tf(x) Tg(y) Tg(z) ]. x=x= y=y= z=z= 1 011 101011000010 0 001 101000010100100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Tweak: W.p., make a random “column” all = x 1. Invariance: Tf, Tg have no matching influential vbl. ⇒ can change dist. to anything w/ same 2-wise corr’s. x=x= y=y= z=z= 1 011 101011000010 0 001 101000010100100 f ( g (g ( g (g (

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Our Dictator Test using NTW Blocks: x 1, y 1,y 2,…,y D unif. random, z i = x 1 y i 1. Then rerandomize x 1. Tweak: W.p., make a random “column” all = x 1. E[ Tf(x) Tg(y) Tg(z) ] ≈ E[ Tf(x) ] E[ Tg(y) Tg(z) ] = 0. x=x= y=y= z=z= 0 001 100011000010 0 001 100000010100100 f ( g (g ( g (g (

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Take-home message If you can design a Dictator Test that seems to work… … you can make it work.

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Open technical problems 1.We need 1 vs. 2 −2 O(D 2 ) hardness for D-to-1. Probably could get away with 2 −poly(D). With 1/poly(D)? 2.Use D-to-1 for other problems. Max-NAE 3 with perfect completeness?

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