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Optimal Inapproximability of Max CSPs over large alphabet Pasin Manurangsi 1 Preetum Nakkiran 2 Luca Trevisan 1 1 UC Berkeley 2 Harvard RANDOM-APPROX 2016 1/17 Max k -CSP R Maximum Constraint Satisfaction Problem: Variables take values in

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SLIDE 1

Optimal Inapproximability of Max CSPs over large alphabet

Pasin Manurangsi1 Preetum Nakkiran2 Luca Trevisan1

1UC Berkeley 2Harvard

RANDOM-APPROX 2016

1/17

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SLIDE 2

Max k-CSPR

Maximum Constraint Satisfaction Problem:

◮ Variables take values in alphabet of size R. ◮ Constraints involve k variables each. ◮ Goal: find assignment maximizing # of satisfied constraints.

2/17

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SLIDE 3

Max k-CSPR

Maximum Constraint Satisfaction Problem:

◮ Variables take values in alphabet of size R. ◮ Constraints involve k variables each. ◮ Goal: find assignment maximizing # of satisfied constraints.

Example

For k = 2, R = 3, a 2-CSP3 is given by a list of constraints:

      

(x1 = 0 ∧ x2 = 2) (x1 = 1 ∧ x3 = 2) . . .

2/17

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SLIDE 4

Hardness of Max k-CSPR

NP-hard to solve exactly (contains MAX-CUT, MAX 3-SAT).

3/17

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SLIDE 5

Hardness of Max k-CSPR

NP-hard to solve exactly (contains MAX-CUT, MAX 3-SAT). NP-hard to approximate (PCP theorem).

3/17

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SLIDE 6

Hardness of Max k-CSPR

NP-hard to solve exactly (contains MAX-CUT, MAX 3-SAT). NP-hard to approximate (PCP theorem). Boolean CSPs (R = 2): Optimal approximation factor is O(k/2k).

3/17

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SLIDE 7

Hardness of Max k-CSPR

NP-hard to solve exactly (contains MAX-CUT, MAX 3-SAT). NP-hard to approximate (PCP theorem). Boolean CSPs (R = 2): Optimal approximation factor is O(k/2k). Non-boolean CSPs (R > 2): not resolved prior.

3/17

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SLIDE 8

Hardness of Approximation

Trivial (1/Rk)-approximation for Max k-CSPR: Random

  • assignment. Each clause matches the maximizing assignment w.p.

1/Rk. Q: Can we do better? Is it hard to do much better?

4/17

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SLIDE 9

Prior Work: Non-boolean Max CSP

Approximation factors:1 Algorithm UG-Hardness NP-Hardness k = 2

log R R

k = 3 3 ≤ k < O(1)

1Ignoring constants, and for large R. 5/17

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SLIDE 10

Prior Work: Non-boolean Max CSP

Approximation factors:1 Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3 3 ≤ k < O(1)

1Ignoring constants, and for large R. 5/17

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SLIDE 11

Prior Work: Non-boolean Max CSP

Approximation factors:1 Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2

3 ≤ k < O(1)

1Ignoring constants, and for large R. 5/17

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SLIDE 12

Prior Work: Non-boolean Max CSP

Approximation factors:1 Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 1 R

3 ≤ k < O(1)

1Ignoring constants, and for large R. 5/17

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SLIDE 13

Prior Work: Non-boolean Max CSP

Approximation factors:1 Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 1 R

3 ≤ k < O(1)

1 Rk−1 1 Rk−2

1Ignoring constants, and for large R. 5/17

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SLIDE 14

Prior Work: Non-boolean Max CSP

Approximation factors:1 Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 1 R

3 ≤ k < O(1)

1 Rk−1 1 Rk−2

For constant k ≥ 3, factor of R gap in hardness vs. approximation.

1Ignoring constants, and for large R. 5/17

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SLIDE 15

Our results

Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 1 R

3 ≤ k < O(1)

1 Rk−1 1 Rk−2

2 6/17

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SLIDE 16

Our results

Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 log R R2 1 R

3 ≤ k < O(1)

1 Rk−1 log R Rk−1 1 Rk−2

2 6/17

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SLIDE 17

Our results

Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 log R R2 log R R2 1 R

3 ≤ k < O(1)

1 Rk−1 log R Rk−1 log R Rk−1 1 Rk−2

2 6/17

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SLIDE 18

Our results

Algorithm UG-Hardness NP-Hardness k = 2

log R R log R R log R √ R

k = 3

1 R2 log R R2 log R R2 1 R

3 ≤ k < O(1)

1 Rk−1 log R Rk−1 log R Rk−1 1 Rk−2

We give matching UG-hardness and approximation algorithms for any k, R. Gap reduced to O(1) for constant k.2

2 Original paper had polylog(R) gap. Improvement suggested by Rishi

Saket, Subhash Khot, Venkat Guriswami.

6/17

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SLIDE 19

Dictator Testing

UG-Hardness-of-approximation equivalent to dictator testing. Dictator: f (x1, x2, . . . , xn) = xi.

7/17

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Dictator Testing

UG-Hardness-of-approximation equivalent to dictator testing. Dictator: f (x1, x2, . . . , xn) = xi.

Problem

Given oracle access to f : [R]n → [R], determine if f is a dictator

  • r “far from a dictator”.

7/17

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SLIDE 21

Dictator Testing

UG-Hardness-of-approximation equivalent to dictator testing. Dictator: f (x1, x2, . . . , xn) = xi.

Problem

Given oracle access to f : [R]n → [R], determine if f is a dictator

  • r “far from a dictator”.

◮ Completeness c: If f is a dictator, accept w.p. ≥ c. ◮ Soundness s: If f is “far from” a dictator, accept w.p. ≤ s.

“Far from dictator” ≡ small low-degree influences (Fourier condition)

7/17

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SLIDE 22

Examples

f : [R]n → R “Far from dictator” ≡ small low-degree influences

Example

Plurality on n coordinates is far from a dictator (no influential coordinate).

8/17

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SLIDE 23

Examples

f : [R]n → R “Far from dictator” ≡ small low-degree influences

Example

Plurality on n coordinates is far from a dictator (no influential coordinate).

Example

f (x1, x2, . . . xn) := x1 ⊕R x2 is NOT far from a dictator.

8/17

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SLIDE 24

UG-Hardness of Approximation

k-query dictator test over alphabet R, with

(soundness, completeness) = (s, c)

⇐ ⇒ UG-hard to distinguish between k-CSPR instances where OPT ≈ s vs. OPT ≈ c

9/17

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SLIDE 25

UG-Hardness of Approximation

k-query dictator test over alphabet R, with

(soundness, completeness) = (s, c)

⇐ ⇒ UG-hard to distinguish between k-CSPR instances where OPT ≈ s vs. OPT ≈ c

  • UG-hard to approximate

Max k-CSPR better than ≈ (s/c).

9/17

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SLIDE 26

UG-hardness of boolean 2-CSP

[Khot, Kindler, Mossel, O’Donnell]

2-Query Boolean Dictator test

f : {0, 1}n → {0, 1}, E[f ] = 1/2.

10/17

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SLIDE 27

UG-hardness of boolean 2-CSP

[Khot, Kindler, Mossel, O’Donnell]

2-Query Boolean Dictator test

f : {0, 1}n → {0, 1}, E[f ] = 1/2.

◮ Pick x ∼ {0, 1}n uniform

10/17

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SLIDE 28

UG-hardness of boolean 2-CSP

[Khot, Kindler, Mossel, O’Donnell]

2-Query Boolean Dictator test

f : {0, 1}n → {0, 1}, E[f ] = 1/2.

◮ Pick x ∼ {0, 1}n uniform ◮ Pick “noise” η ∼ {0, 1}n, each coordinate Bernoulli(p).

10/17

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SLIDE 29

UG-hardness of boolean 2-CSP

[Khot, Kindler, Mossel, O’Donnell]

2-Query Boolean Dictator test

f : {0, 1}n → {0, 1}, E[f ] = 1/2.

◮ Pick x ∼ {0, 1}n uniform ◮ Pick “noise” η ∼ {0, 1}n, each coordinate Bernoulli(p). ◮ Accept iff f (x) = f (x ⊕ η)

10/17

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SLIDE 30

UG-hardness of boolean 2-CSP

[Khot, Kindler, Mossel, O’Donnell]

2-Query Boolean Dictator test

f : {0, 1}n → {0, 1}, E[f ] = 1/2.

◮ Pick x ∼ {0, 1}n uniform ◮ Pick “noise” η ∼ {0, 1}n, each coordinate Bernoulli(p). ◮ Accept iff f (x) = f (x ⊕ η)

For p ≈ 0.15,

◮ Completeness: If f is a dictator, accepts w.p.

≥ 1 − p ≈ 0.85.

◮ Soundness: If f is “far from” a dictator, accepts w.p.

≤≈ 0.74.

10/17

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SLIDE 31

UG-hardness of boolean 2-CSP

[Khot, Kindler, Mossel, O’Donnell]

2-Query Boolean Dictator test

f : {0, 1}n → {0, 1}, E[f ] = 1/2.

◮ Pick x ∼ {0, 1}n uniform ◮ Pick “noise” η ∼ {0, 1}n, each coordinate Bernoulli(p). ◮ Accept iff f (x) = f (x ⊕ η)

For p ≈ 0.15,

◮ Completeness: If f is a dictator, accepts w.p.

≥ 1 − p ≈ 0.85.

◮ Soundness: If f is “far from” a dictator, accepts w.p.

≤≈ 0.74.

◮ Ratio: s/c ≈ 0.878567 = αGW

10/17

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SLIDE 32

Why it works

Verifier accepts iff f (x) = f (x + η) Noise η iid on every coordinate.

11/17

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SLIDE 33

Why it works

Verifier accepts iff f (x) = f (x + η) Noise η iid on every coordinate. If f depends on many coordinates, the noise will “add up”: f (x + η) will be almost uncorrelated with f (x).

11/17

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SLIDE 34

Why it works

Verifier accepts iff f (x) = f (x + η) Noise η iid on every coordinate. If f depends on many coordinates, the noise will “add up”: f (x + η) will be almost uncorrelated with f (x).

Example

majority function maj : {±1}n → {±1}. maj(x1, . . . , xn) = sign(

  • i

xi) If noise η is high enough, sign(

i xi) will be almost independent of

sign(

i(xi + ηi))

11/17

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SLIDE 35

Our k-query large alphabet dictator test

f : [R]n → [R] f is balanced: All pre-images f −1(i) of same size.

12/17

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SLIDE 36

Our k-query large alphabet dictator test

f : [R]n → [R] f is balanced: All pre-images f −1(i) of same size.

◮ Pick z ∼ [R]n uniform

12/17

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SLIDE 37

Our k-query large alphabet dictator test

f : [R]n → [R] f is balanced: All pre-images f −1(i) of same size.

◮ Pick z ∼ [R]n uniform ◮ Pick k iid noise η1, . . . , ηk ∈ [R]n, s.t. each coordinate of ηj is

  • w.p. ρ

uniform in [R]

  • therwise

12/17

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SLIDE 38

Our k-query large alphabet dictator test

f : [R]n → [R] f is balanced: All pre-images f −1(i) of same size.

◮ Pick z ∼ [R]n uniform ◮ Pick k iid noise η1, . . . , ηk ∈ [R]n, s.t. each coordinate of ηj is

  • w.p. ρ

uniform in [R]

  • therwise

◮ Accept iff f (z + η1) = f (z + η2) = · · · = f (z + ηk)

12/17

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SLIDE 39

Our Results

f : [R]n → [R] Accept iff f (z + η1) = f (z + η2) = · · · = f (z + ηk)

13/17

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SLIDE 40

Our Results

f : [R]n → [R] Accept iff f (z + η1) = f (z + η2) = · · · = f (z + ηk) We show:

◮ Completeness: If f is a dictator (f (x) = xj), accepts

w.p. ≈

1 (log R)k/2 ◮ Soundness: If f is balanced and has small influences, accepts

w.p. ≤≈

1 Rk−1

If f is far from dictator, the k queries f (z + η1), f (z + η2), . . . look almost independent – all equal w.p. ≈

1 Rk−1 .

13/17

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SLIDE 41

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i]

14/17

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SLIDE 42

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced.

14/17

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SLIDE 43

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced. Pr[accept] = Pr[f (x + η1) = f (x + η2) = · · · = f (x + ηk)]

14/17

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SLIDE 44

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced. Pr[accept] = Pr[f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

Pr[i = f (x + η1) = f (x + η2) = · · · = f (x + ηk)]

14/17

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SLIDE 45

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced. Pr[accept] = Pr[f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

Pr[i = f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

E

x,η[f i(x + η1)f i(x + η2) . . . f i(x + ηk)]

14/17

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SLIDE 46

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced. Pr[accept] = Pr[f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

Pr[i = f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

E

x,η[f i(x + η1)f i(x + η2) . . . f i(x + ηk)]

Define gi(x) = Eη[f i(x + η)] (i.e. gi := Tρf i)

14/17

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SLIDE 47

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced. Pr[accept] = Pr[f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

Pr[i = f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

E

x,η[f i(x + η1)f i(x + η2) . . . f i(x + ηk)]

=

  • i∈[R]

E

x [(gi(x))k]

Define gi(x) = Eη[f i(x + η)] (i.e. gi := Tρf i)

14/17

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SLIDE 48

Soundness Analysis Ideas

Define f i : [R]n → {0, 1} as f i(x) := ✶[f (x) = i] E[f i] = 1/R since f is balanced. Pr[accept] = Pr[f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

Pr[i = f (x + η1) = f (x + η2) = · · · = f (x + ηk)] =

  • i∈[R]

E

x,η[f i(x + η1)f i(x + η2) . . . f i(x + ηk)]

=

  • i∈[R]

E

x [(gi(x))k]

(want) ≈

  • i∈[R]

E

x [(gi(x))]k =

  • i

(1/R)k = 1/Rk−1 Define gi(x) = Eη[f i(x + η)] (i.e. gi := Tρf i)

14/17

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SLIDE 49

Soundness Analysis Ideas

f : [R]n → [0, 1], E[f ] = 1/R

15/17

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SLIDE 50

Soundness Analysis Ideas

f : [R]n → [0, 1], E[f ] = 1/R g(x) = E

η [f (x + η)] = (Tρf )(x)

15/17

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SLIDE 51

Soundness Analysis Ideas

f : [R]n → [0, 1], E[f ] = 1/R g(x) = E

η [f (x + η)] = (Tρf )(x)

Want to show: E[(Tρf )k] E[f ]k ⇐ ⇒ ||Tρf ||k ||f ||1

15/17

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SLIDE 52

Soundness Analysis Ideas

f : [R]n → [0, 1], E[f ] = 1/R g(x) = E

η [f (x + η)] = (Tρf )(x)

Want to show: E[(Tρf )k] E[f ]k ⇐ ⇒ ||Tρf ||k ||f ||1 This is hypercontractivity

15/17

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SLIDE 53

Soundness Analysis Ideas

f : [R]n → [0, 1], E[f ] = 1/R g(x) = E

η [f (x + η)] = (Tρf )(x)

Want to show: E[(Tρf )k] E[f ]k ⇐ ⇒ ||Tρf ||k ||f ||1 This is hypercontractivity

15/17

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SLIDE 54

Algorithm Ideas

k-CSPR := k variables per clause; variables over domain [R]

16/17

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Algorithm Ideas

k-CSPR := k variables per clause; variables over domain [R]

k = 2 :

Existing ( log R

R )-approximation algo (SDP-based).

Advantage of A = R log R over random assignment ( 1

R2 ).

16/17

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SLIDE 56

Algorithm Ideas

k-CSPR := k variables per clause; variables over domain [R]

k = 2 :

Existing ( log R

R )-approximation algo (SDP-based).

Advantage of A = R log R over random assignment ( 1

R2 ).

k ≥ 3 :

Reduction to k = 2 case, preserving the Advantage over random assignment. = ⇒ ( log R

Rk−1 )-approximation algo for k ≥ 3.

16/17

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SLIDE 57

Algorithm Ideas

k-CSPR := k variables per clause; variables over domain [R]

k = 2 :

Existing ( log R

R )-approximation algo (SDP-based).

Advantage of A = R log R over random assignment ( 1

R2 ).

k ≥ 3 :

Reduction to k = 2 case, preserving the Advantage over random assignment. = ⇒ ( log R

Rk−1 )-approximation algo for k ≥ 3.

Map a constraint (X1 = a1) ∧ (X2 = a2) ∧ · · · ∧ (Xk = ak) to all pairwise constraints {(Xi = ai ∧ Xj = aj) : 1 ≤ i < j ≤ k}

16/17

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SLIDE 58

Conclusions

We know the approximability of Max k-CSPR for constant k, assuming Unique Games Conjecture: Θ( log R

Rk−1 )

17/17

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SLIDE 59

Conclusions

We know the approximability of Max k-CSPR for constant k, assuming Unique Games Conjecture: Θ( log R

Rk−1 )

Previously we knew it was between

1 Rk−2 and 1 Rk−1 (NP-hardness).

17/17

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SLIDE 60

Conclusions

We know the approximability of Max k-CSPR for constant k, assuming Unique Games Conjecture: Θ( log R

Rk−1 )

Previously we knew it was between

1 Rk−2 and 1 Rk−1 (NP-hardness).

Open:

◮ Tight results for larger k < R. (already known for k ≥ R).

Current gap is log R

Rk−1 vs. k2 log(kR) Rk−1

17/17

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SLIDE 61

Conclusions

We know the approximability of Max k-CSPR for constant k, assuming Unique Games Conjecture: Θ( log R

Rk−1 )

Previously we knew it was between

1 Rk−2 and 1 Rk−1 (NP-hardness).

Open:

◮ Tight results for larger k < R. (already known for k ≥ R).

Current gap is log R

Rk−1 vs. k2 log(kR) Rk−1 ◮ Tight results based on NP-hardness

17/17