Social Choice,

Computational Complexity,

Gaussian Geometry,

& Boolean Functions and

Ryan O’Donnell

Carnegie Mellon University

analysisofbooleanfunctions.org

f : {−1,+1}n → {−1,+1}{−1,+1}n

+1

+1

+1+1

−1

−1

−1−1

f = ±1S

S ⊆ {−1,+1}n

Form “ρ-correlated”

For each 1 ≤ i ≤ n…

with probability ρ,

with probability 1−ρ,

+

+

−

−

+

+

+

+

−

+

−

−

+

+

−

−

−

−

+1

+1

+1+1

−1

−1

−1−1

f : {−1,+1}n → {−1,+1}{−1,+1}n

For .9-correlated

x

f = ±1S

ρ-Sensitivity[f] :=

a kind of measure of the “boundary size” of S

We’ll focus on “volume - sets” S.

Equivalently, “balanced” f:

Which balanced f : {−1,+1}n → {−1,+1}

minimizes ρ-Sensitivity[f]?

ρ-Isoperimetric Problem on Discrete Cube

Social Choice interpretation

Election with n voters, 2 candidates named ±1.

f : {−1,+1}n → {−1,+1} is the voting rule:

xj ∈ {−1,+1} is jth voter’s preference.

f(x) = f(x1, …, xn) = winner of the election.

E.g.: f(x) = Majority(x) = sgn(x1 + ∙∙∙ + xn)

f(x) = ElectoralCollege(x)

f(x) = +1 (not balanced)

Social Choice interpretation

Impartial Culture Assumption [GK’68]:

Voters’ preferences are uniformly random.

“Faulty voting machine twist”:

Each vote recorded correctly with prob. ρ,

changed to a random vote with prob. 1−ρ.

ρ-Sens[f] = Pr[faulty machines affect outcome]

Which balanced f : {−1,+1}n → {−1,+1}

minimizes ρ-Sensitivity[f]?

Answer:

Dictatorships, f(x) = xj

(and negated-dictatorships, f(x) = −xj)

+1

+1

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−1

−1

−1

+1

+1

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−1

−1

−1

+1+1

+1+1

−1

−1 −1

−1

−1−1

−1−1

+1

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+1

Which balanced f : {−1,+1}n → {−1,+1}

minimizes ρ-Sensitivity[f]?

Theorem:

∀ balanced f : {−1,+1}n → {−1,+1},

ρ-Sens[f] ≥ ρ-Sens[±Dictators] = (1−ρ)/2

Proof:

Fourier analysis of Boolean functions.

One more social choice detour…

Three candidates A, B, C, ranked by n voters.

Societal ranking produced by holding 3 pairwise

elections using some f : {−1,+1}n → {−1,+1}. (Condorcet election / Independence of Irrelevant Alternatives)

Condorcet’s Paradox (1785): With f = Majority,

might obtain “A beats B, B beats C, C beats A”!

Arrow’s Theorem (1950):

Paradox never occurs ⇒ f = ±Dictator. ☹

Kalai’s Proof (2002):

Same Fourier analysis as in previous theorem.

Every mathematics talk should contain…

a joke

a proof

+1

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−1−1

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j

Infj [ ith Dictator ]

=

Examples:

Infj [Majorityn] ∀ j

Which balanced f : {−1,+1}n → {−1,+1}

with Influencej[f] “small” for all 1 ≤ j ≤ n

minimizes ρ-Sensitivity[f]?

Stablest voting rule problem

If f : {−1,+1}n → {−1,+1} is balanced,

and Influencej[f] ≤ δ for all 1 ≤ j ≤ n, then

ρ-Sens[f] ≥ ρ-Sens[Majority] − ϵ(δ)

(where ϵ(δ) → 0 as δ → 0)

Majority Is Stablest Conjecture [KKMO’04]

[Guilbaud’52]

If f : {−1,+1}n → {−1,+1} is balanced,

and Influencej[f] ≤ δ for all 1 ≤ j ≤ n, then

ρ-Sens[f] ≥ ρ-Sens[Majority] − ϵ(δ)

(where ϵ(δ) → 0 as δ → 0)

Majority Is Stablest

[Guilbaud’52]

Theorem [MOO’05]

ρ 10

0

ρ-Sens

(1−ρ)/2

(quality of voting machines)

(pro

babili

ty o

utc

om

e a

ffect

ed

)

If f : {−1,+1}n → {−1,+1} is balanced,

and Influencej[f] ≤ δ for all 1 ≤ j ≤ n, then

ρ-Sens[f] ≥ ρ-Sens[Majority] − ϵ(δ)

(where ϵ(δ) → 0 as δ → 0)

Majority Is Stablest [KKMO’04/MOO’05]

2013: New proof by De, Mossel, Neeman

[KKMO’04] motivation:

“Majority Is Stablest” is the exact statement

needed to show an optimal computational

complexity result for the algorithmic task called

Maximum-Cut.

Max-Cut

Input:

“Almost bipartite” N-vertex graph

Output:

Optimal bipartition

“mistake edges”

Max-Cut

“Brute force” algorithm: ≈ 2N steps.

Question:

Is there an “efficient” (= NC step) algorithm?

Answer:

No. (Assuming “P≠NP”. Max-Cut is “NP-hard”.)

Max-Cut

Input:

“Almost bipartite” N-vertex graph

Output:

“mistake edges”

ApproximateOptimal bipartitionDo your best

There is an efficient algorithm s.t. ∀ ρ

if input graph is “ρ-bipartite”,

then algorithm outputs a bipartition

with fraction of mistake edges ≤

Theorem: [GLS’88,DP’90,GW’94]

“optimal bipartition has ≤ (1−ρ)/2 fraction of mistake edges”

≥ .69

ρ 10

0

(1−ρ)/2

How bipartite the input graph is

Fraction ofmistake edges

GW alg’sguarantee

optimalbipartition

prev bestefficient

algorithm

[KKMO’04] Theorem:

“Majority Is Stablest”

⇒ NP-hard to beat GW’s Max-Cut algorithm

“UG-hard”

Raghavendra ’08: (see also [KKMO’04,Aus’06,Aus’07,OW’07,RS08])

∃ a generic, efficient algorithm A such that

for all “constraint satisfaction problems” M,

it’s UG-hard to approx. M better than A does.

Proving Majority Is Stablest:

enter Gaussian geometry.

+1

+1

+1+1

−1

−1

−1−1

f : {−1,+1}n → {−1,+1} balanced{−1,+1}n

For ρ-correlated

x

f = ±1S

If f : {−1,+1}n → {−1,+1} is balanced,

and Influencej[f] ≤ δ for all 1 ≤ j ≤ n, then

ρ-Sens[f] ≥ “ρ-Sens[Majority]” − ϵ(δ)

(where ϵ(δ) → 0 as δ → 0)

Majority Is Stablest Theorem

[exercise, Sheppard 1899]

“Gaussian-ρ-Sensitivity”[sgn]

sgn : ℝ1 → {−1,+1}

ℝ1

= ±1S

(Note: S has Gaussian volume ½; i.e., sgn is “balanced”.)

n-dim. Boolean function Majority is the 1-dim. Gaussian function sgn in disguise!

z

S = (0,∞) ⊆ ℝ1

More generally, for g : ℝd → {−1,+1}, g = ±1S, define

Gaussian-ρ-Sens[g] =

The Gaussian function g can be “disguised” bya sequence of (small-influence) Boolean functions.

S

ℝ2

As n → ∞: • ρ-Sens[f] → Gaussian-ρ-Sens[g]

• if g is “balanced” (Pr [z ∈ S] = ½),

f → balanced

• Influencej[f] → 0 ∀ j

Majority Is Stablesthypotheses

If g : ℝd → {−1,+1} is balanced,

Gaussian-ρ-Sens[g] ≥ .

∴Majority Is Stablest Theorem implies…

Gaussian-ρ-Sens [sgn]

Borell’s Isoperimetric Inequality

∴Majority Is Stablest Theorem implies…

[Borell ’85](special case)

Equality if S is halfline in ℝ1, orindeed any halfspace thru 0 in ℝd

If S ⊆ ℝd has Gaussian volume ½,

ρ → 1 implies classical Gaussian Isoperimetric Inequality[Borell’74, Sudakov−Tsirelson’74]

∴Majority Is Stablest ⇒ Borell’s Isoperim. Ineq.

Proofs of Borell’s Isoperimetric Inequality:

• Borell ’85: Gaussian rearrangement, very hard• Beckner ’90:

Analogue on the sphere by 2-point symm.,pretty easy, implies Gaussian version [CL’90]

• [KO’12]: vol.-½, : four sentences

Every mathematics talk should contain…

a joke

a proof

∴Majority Is Stablest ⇒ Borell’s Isoperim. Ineq.

Proofs of Borell’s Isoperimetric Inequality:

• Borell ’85: Gaussian rearrangement, very hard• Beckner ’90:

Analogue on the sphere by 2-point symm.,pretty easy, implies Gaussian version [CL’90]

• [KO’12]: vol.-½, : four sentences

First proof of Majority Is Stablest:

[MOO’05] proved “Invariance Principle” (nonlinear CLT)

to obtain Borell’s Isoperim. Ineq. ⇒ Majority Is Stablest,

whence UG-hardness of beating GW Max-Cut algorithm.

∴Majority Is Stablest ⇒ Borell’s Isoperim. Ineq.

Proofs of Borell’s Isoperimetric Inequality:

• Borell ’85: Gaussian rearrangement, very hard• Beckner ’90:

Analogue on the sphere by 2-point symm.,pretty easy, implies Gaussian version [CL’90]

• [MN’12]: Semigroup method

• [DMN’13]: Discrete proof of Majority Is Stablest (hence also Borell’s Isoperimetric Ineq.) by induction on n.

• [KO’12]: vol.-½, : four sentences

• Eldan ’13: Stochastic calculus

Conclusion: Importance of multiple proofs

[MOO] proof of Majority Is Stablest:

• Invariance Principle, reduced to Gaussian geom.

• Advantage: Invariance Principle useful elsewhere: Social Choice, Learning Theory, Comp. Complexity [Raghavendra’08]

[DMN] proof of Majority Is Stablest:

• Direct induction on n, completely discrete

• Advantage: Proof expressible in “SOS proof system”, which has algorithmic implications…

Thanks!