Fourier Analysis, Projections, Influence, Junta, Etc…

© S.Safra Boolean Functions and Juntas Def: A Boolean function

© S.Safra f f * * -1* 1* 11* 11-1* -1-1* -11* -11-1* -111* * -1-11* 111* 1-1* 1-1-1* 1-11* Functions as an Inner-Product Vector-Space f f 2n2n 2n2n * * -1* 1* 11* 11-1* -1-1* -11* -11-1* -111* * -1-11* 111* 1-1* 1-1-1* 1-11*

© S.Safra Functions’ Vector-Space A functions f is a vector A functions f is a vector Addition: ‘f+g’(x) = f(x) + g(x) Addition: ‘f+g’(x) = f(x) + g(x) Multiplication by scalar ‘c  f’(x) = c  f(x) Multiplication by scalar ‘c  f’(x) = c  f(x)

© S.Safra Variables` Influence The influence of an index i  [n] on a Boolean function f:{1,-1} n  {1,-1} is The influence of an index i  [n] on a Boolean function f:{1,-1} n  {1,-1} is

© S.Safra Norms Def: Expectation Norm Def: Expectation Norm Def: Sum Norm Def: Sum Norm

© S.Safra Inner-Product A functions f is a vector A functions f is a vector Inner product (normalized) Inner product (normalized)

© S.Safra Simple Observations Claims: For any function f whose range is {-1, 0, 1}

© S.Safra Monomials What would be the monomials over x  P[n]? What would be the monomials over x  P[n]? All powers except 0 and 1 disappear! All powers except 0 and 1 disappear! Hence, one for each character S  [n] Hence, one for each character S  [n] These are all the multiplicative functions These are all the multiplicative functions

© S.Safra Fourier-Walsh Transform Consider all characters Consider all characters Given any function let the Fourier-Walsh coefficients of f be Given any function let the Fourier-Walsh coefficients of f be thus f can described as thus f can described as

© S.Safra Fourier Transform: Norm Norm: (Sum) Thm [Parseval]: Hence, for a Boolean f

© S.Safra Variables` Influence The influence of an index i  [n] on a Boolean function f:{1,-1} n  {1,-1} is The influence of an index i  [n] on a Boolean function f:{1,-1} n  {1,-1} is Which can be expressed in terms of the Fourier coefficients of f Claim: Which can be expressed in terms of the Fourier coefficients of f Claim:

© S.Safra Restriction and Average Def: Let I  [n], x  P([n]\I), the restriction function is Def: the average function is Note: [n] I x y I x y y y y y

© S.Safra In Fourier Expansion Prop: Prop: And since the expectation of a function is its coefficient on the empty character: And since the expectation of a function is its coefficient on the empty character: Corollary: Corollary:

© S.Safra Expectation and Variance Claim: Claim: Hence, for any f Hence, for any f

© S.Safra Average Sensitivity Def: the sensitivity of x w.r.t. f is Def: the sensitivity of x w.r.t. f is Def: the average-sensitivity of f is Def: the average-sensitivity of f is

© S.Safra When as(f)=1 Def: a balanced function f is s.t. Def: a balanced function f is s.t. Thm: a balanced, Boolean f s.t. as(f)=1 is a dictatorship Thm: a balanced, Boolean f s.t. as(f)=1 is a dictatorship Proof: observe that and since as(f)=1 it must be that however ||f|| 2 =1 hence Proof: observe that and since as(f)=1 it must be that however ||f|| 2 =1 hence

© S.Safra Linear, Boolean Functions Proof(cont): Proof(cont): pick any x; f(x)  {-1, 1} pick any x; f(x)  {-1, 1} Pick {i} with non-zero coefficient Pick {i} with non-zero coefficient Observe that f(x  {i})  {-1, 1} however differ from f(x) Observe that f(x  {i})  {-1, 1} however differ from f(x) Conclusion: Conclusion:

© S.Safra Codes and Boolean Functions Def: an m-bit code is a subset of the set of all the m- binary string C  {-1,1} m The distance of a code C, which is the minimum, over all pairs of legal-words (in C), of the Hamming distance between the two words A Boolean function over n binary variables, is a 2 n -bit string Hence, a set of Boolean functions can be considered as a 2 n -bits code

© S.Safra Hadamard Code In the Hadamard code the set of legal-words consists of all multiplicative (linear if over {0,1}) functions C={  S | S  [n]} namely all characters

© S.Safra Hadamard Test Given a Boolean f, choose random x and y; check that f(x)f(y)=f(xy) Prop(completeness): a legal Hadamard word (a character) always passes this test

22 Hadamard Test – Soundness Prop(soundness): Proof:

© S.Safra Long-Code In the long-code the set of legal-words consists of all monotone dictatorships This is the most extensive binary code, as its bits represent all possible binary values over n elements

© S.Safra Long-Code Encoding an element e  [n] : Encoding an element e  [n] : E e legally-encodes an element e if E e = f e E e legally-encodes an element e if E e = f e F F F F T T T T T T

© S.Safra Testing Long-code Def(a long-code list-test): given a code-word f, probe it in a constant number of entries, and accept almost always if f is a monotone dictatorship accept almost always if f is a monotone dictatorship reject w.h.p if f does not have a sizeable fraction of its Fourier weight concentrated on a small set of variables, that is, if  a semi-Junta J  [n] s.t. reject w.h.p if f does not have a sizeable fraction of its Fourier weight concentrated on a small set of variables, that is, if  a semi-Junta J  [n] s.t. Note: a long-code list-test, distinguishes between the case f is a dictatorship, to the case f is far from a junta.

© S.Safra Motivation – Testing Long-code The long-code list-test are essential tools in proving hardness results. The long-code list-test are essential tools in proving hardness results. Hence finding simple sufficient-conditions for a function to be a junta is important. Hence finding simple sufficient-conditions for a function to be a junta is important.

© S.Safra Perturbation Def: denote by   the distribution over all subsets of [n], which assigns probability to a subset x as follows: independently, for each i  [n], let i  x with probability 1-  i  x with probability 1-  i  x with probability  i  x with probability 

© S.Safra Long-Code Test Given a Boolean f, choose random x and y, and choose z   ; check that f(x)f(y)=f(xyz) Prop(completeness): a legal long- code word (a dictatorship) passes this test w.p. 1- 

29 Long-code Test – Soundness Prop(soundness): Proof:

© S.Safra Variation Def: the variation of f (extension of influence ) Prop: the following are equivalent definitions to the variation of f:

© S.Safra Proof Recall Therefore

© S.Safra Proof – Cont. Recall Recall Therefore (by Parseval): Therefore (by Parseval):

© S.Safra High vs Low Frequencies Def: The section of a function f above k is and the low-frequency portion is

© S.Safra Junta Test Def: A Junta test is as follows: A distribution over l queries For each l-tuple, a local-test that either accepts or rejects:T[x 1, …, x l ]: {1, -1} l  {T,F} s.t. for a j-junta f whereas for any f which is not ( , j)-Junta

© S.Safra Fourier Representation of influence Proof: consider the I-average function on P[I] which in Fourier representation is and

© S.Safra Fourier Representation of influence Proof: consider the influence function which in Fourier representation is and

© S.Safra Subsets` Influence Def: The Variation of a subset I  [n] on a Boolean function f is and the low-frequency influence

© S.Safra Independence-Test The I-independence-test on a Boolean function f is, for Lemma:

© S.Safra

Junta Test The junta-size test JT on a Boolean function f is The junta-size test JT on a Boolean function f is Randomly partition [n] to I 1,.., I r Randomly partition [n] to I 1,.., I r Run the independence-test t times on each I h Run the independence-test t times on each I h Accept if ≤j of the I h fail their independence-tests Accept if ≤j of the I h fail their independence-tests For r>>j 2 and t>>j 2 / 

© S.Safra Completeness Lemma: for a j-junta f Proof: only those sets which contain an index of the Junta would fail the independence-test

© S.Safra Soundness Lemma: Proof: Assume the premise. Fix  <<1/t and let

© S.Safra |J| ≤ j Prop: r >> j implies |J| ≤ j Proof: otherwise, J spreads among I h w.h.p. J spreads among I h w.h.p. and for any I h s.t. I h  J ≠  it must be that Variation I h (f) >  and for any I h s.t. I h  J ≠  it must be that Variation I h (f) > 

© S.Safra High Frequencies Contribute Little Prop: k >> r log r implies Proof: a character S of size larger than k spreads w.h.p. over all parts I h, hence contributes to the influence of all parts. If such characters were heavy (>  /4), then surely there would be more than j parts I h that fail the t independence-tests

© S.Safra Almost all Weight is on J Lemma: Proof: assume by way of contradiction otherwise since for a random partition w.h.p. (Chernoff bound) for every h however, since for any I the influence of every I h would be ≥  /100rk

© S.Safra Find the Close Junta Now, since consider the (non Boolean) which, if rounded outside J is Boolean and not more than  far from f

© S.Safra Consider the q-biased product distribution  q : Def: The probability of a subset F and for a family of subsets  Consider the q-biased product distribution  q : Def: The probability of a subset F and for a family of subsets  Product, Biased Distribution

© S.Safra Beckner/Nelson/Bonami Inequality Def: let T  be the following operator on any f, Prop: Proof:

© S.Safra Beckner/Nelson/Bonami Inequality Def: let T  be the following operator on any f, Thm: for any p≥r and  ≤((r-1)/(p-1)) ½

© S.Safra Beckner/Nelson/Bonami Corollary Corollary 1: for any real f and 2≥r≥1 Corollary 2: for real f and r>2

© S.Safra Average Sensitivity The sum of variables’ influence is referred to as the average sensitivity Which can be expressed by the Fourier coefficients as

© S.Safra Freidgut Theorem Thm: any Boolean f is an [ , j]-junta for Proof: 1. Specify the junta J 2. Show the complement of J has little influence

© S.Safra Specify the Junta Set k=  (as(f)/  ), and  =2 -  (k) Let We’ll prove: and let hence, J is a [ ,j]-junta, and |J|=2 O(k)

© S.Safra High Frequencies Contribute Little Prop: Proof: a character S of size larger than k contributes at least k times the square of its coefficient to the average sensitivity. If such characters were heavy (>  /4), as(f) would have been large

© S.Safra Altogether Lemma: Proof:

© S.Safra Altogether

Biased  q - Influence The  q -influence of an index i  [n] on a boolean function f:P[n]  {1,-1} is The  q -influence of an index i  [n] on a boolean function f:P[n]  {1,-1} is

© S.Safra Biased Walsh Product The usual Fourier basis  is not orthogonal with respect to the biased inner-product, The usual Fourier basis  is not orthogonal with respect to the biased inner-product, Hence, we use the Biased Walsh Product: Hence, we use the Biased Walsh Product:

© S.Safra Thm [Margulis-Russo]: For monotone f Hence Lemma: For monotone f  > 0,  q  [p, p+  ] s.t. as q (f)  1/  Proof: Otherwise  p+  (f) > 1

© S.Safra Proof [Margulis-Russo]:

© S.Safra Influential People and Issues The theory of the influence of variables on Boolean functions [BL, KKL] and related issues, has been introduced to tackle social choice problems, furthermore has motivated a magnificent sequence of works, related to Economics [K], percolation [BKS], Hardness of approximation [DS] Revolving around the Fourier/Walsh analysis of Boolean functions… The theory of the influence of variables on Boolean functions [BL, KKL] and related issues, has been introduced to tackle social choice problems, furthermore has motivated a magnificent sequence of works, related to Economics [K], percolation [BKS], Hardness of approximation [DS] Revolving around the Fourier/Walsh analysis of Boolean functions… And the real important question: And the real important question:

© S.Safra Where to go for Dinner? Who has suggestions: Each cast their vote in an (electronic) envelope, and have the system decided, not necessarily according to majority… It turns out someone –in the Florida wing- has the power to flip some votes Power influence

© S.Safra Voting Systems n agents, each voting either “for” (T) or “against” (F) – a Boolean function over n variables f is the outcome n agents, each voting either “for” (T) or “against” (F) – a Boolean function over n variables f is the outcome The values of the agents (variables) may each, independently, flip with probability The values of the agents (variables) may each, independently, flip with probability It turns out: one cannot design an f that would be robust to such noise -that is, would, on average, change value w.p. < O(1) - unless taking into account only very few of the votes It turns out: one cannot design an f that would be robust to such noise -that is, would, on average, change value w.p. < O(1) - unless taking into account only very few of the votes

© S.Safra [n] x I I z Noise-Sensitivity How often does the value of f changes when the input is perturbed? x I I z

© S.Safra Def( ,p,x [n] ): Let 0< <1, and x  P([n]). Then y~ ,p,x, if y = (x\I)  z where Def( ,p,x [n] ): Let 0< <1, and x  P([n]). Then y~ ,p,x, if y = (x\I)  z where I~  [n] is a noise subset, and I~  [n] is a noise subset, and z~  p I is a replacement. z~  p I is a replacement. Def( -noise-sensitivity): let 0< <1, then [ When p=½ equivalent to flipping each coordinate in x independently w.p. /2.] [n] x I I z Noise-Sensitivity

© S.Safra Noise-Sensitivity – Cont. Advantage: very efficiently testable (using only two queries) by a perturbation-test. Advantage: very efficiently testable (using only two queries) by a perturbation-test. Def (perturbation-test): choose x~  p, and y~ ,p,x, check whether f(x)=f(y) The success is proportional to the noise- sensitivity of f. Def (perturbation-test): choose x~  p, and y~ ,p,x, check whether f(x)=f(y) The success is proportional to the noise- sensitivity of f. Prop: the -noise-sensitivity is given by Prop: the -noise-sensitivity is given by

© S.Safra Relation between Parameters Prop: small ns  small high-freq weight Proof: therefore: if ns is small, then Hence the high frequencies must have small weights (as). Prop: small as  small high-freq weight Proof:

© S.Safra Main Result Theorem:  constant  >0 s.t. any Boolean function f:P([n])  {-1,1} satisfying is an [ ,j]-junta for j=O(  -2 k 3  2k ). Corollary: fix a p-biased distribution  p over P([n]) Let >0 be any parameter. Set k=log 1- (1/2) Then  constant  >0 s.t. any Boolean function f:P([n])  {-1,1} satisfying is an [ ,j]-junta for j=O(  -2 k 3  2k )

© S.Safra First Attempt: Following Freidgut’s Proof Thm: any Boolean function f is an [ ,j]-junta for Proof: 1. Specify the junta where, let k=O(as(f)/  ) and fix  =2 -O(k) 2. Show the complement of J has small variation P([n]) J

© S.Safra If k were 1 Easy case (!?!): If we’d have a bound on the non- linear weight, we should be done. The linear part is a set of independent characters (the singletons) In order for those to hit close to 1 or -1 most of the time, they must avoid the Law of Large Numbers, namely be almost entirely placed on one singleton [by Chernoff like bound] Thm[FKN, ext.]: Assume f is close to linear, then f is close to shallow (  a constant function or a dictatorship)

© S.Safra How to Deal with Dependency between Characters Recall Recall (theorem’s premise) (theorem’s premise) Idea: Let Partition [n]\J into I 1,…,I r, for r >> k Partition [n]\J into I 1,…,I r, for r >> k w.h.p f I [x] is close to linear (low freq characters intersect I expectedly by  1 element, while high-frequency weight is low). w.h.p f I [x] is close to linear (low freq characters intersect I expectedly by  1 element, while high-frequency weight is low). P([n]) J I1I1 I2I2 IrIr I

© S.Safra Shallow Function Def: a function f is linear, if only singletons have non-zero weight Def: a function f is linear, if only singletons have non-zero weight Def: a function f is shallow, if f is either a constant or a dictatorship. Def: a function f is shallow, if f is either a constant or a dictatorship. Claim: Boolean linear functions are shallow. Claim: Boolean linear functions are shallow. 0123kn0123kn weight Character size

© S.Safra Boolean Linear  Shallow Claim: Boolean linear functions are shallow. Claim: Boolean linear functions are shallow. Proof: let f be Boolean linear function, we next show: Proof: let f be Boolean linear function, we next show: 1.  {i o } s.t. (i.e. ) 2. And conclude, that eitheror i.e. f is shallow

© S.Safra Claim 1 Claim 1: let f be Boolean linear function, then  {i o } s.t. Claim 1: let f be Boolean linear function, then  {i o } s.t. Proof: w.l.o.g assume Proof: w.l.o.g assume for any z  {3,…,n} consider x 00 =z, x 10 =z  {1}, x 01 =z  {2}, x 11 =z  {1,2} for any z  {3,…,n} consider x 00 =z, x 10 =z  {1}, x 01 =z  {2}, x 11 =z  {1,2} then. then. Next value must be far from {-1,1} Next value must be far from {-1,1} A contradiction! (boolean function) A contradiction! (boolean function) Therefore Therefore 1 ?

© S.Safra Claim 1 Claim 1: let f be Boolean linear function, then  {i o } s.t. Claim 1: let f be Boolean linear function, then  {i o } s.t. Proof: w.l.o.g assume Proof: w.l.o.g assume for any z  {3,…,n} consider x 00 =z, x 10 =z  {1}, x 01 =z  {2}, x 11 =z  {1,2} for any z  {3,…,n} consider x 00 =z, x 10 =z  {1}, x 01 =z  {2}, x 11 =z  {1,2} then. then. But this is impossible as f(x 00 ),f(x 10 ),f(x 01 ), f(x 11 )  {-1,1}, hence their distances cannot all be >0 ! But this is impossible as f(x 00 ),f(x 10 ),f(x 01 ), f(x 11 )  {-1,1}, hence their distances cannot all be >0 ! Therefore. Therefore. 1 ?

© S.Safra Claim 2 Claim 2: let f be Boolean function, s.t. Then eitheror Claim 2: let f be Boolean function, s.t. Then eitheror Proof: consider f(  ) and f(i 0 ): Proof: consider f(  ) and f(i 0 ): Then Then but f is Boolean, hence but f is Boolean, hence therefore therefore 1 0

© S.Safra Linearity and Dictatorship Prop: Let f be a balanced linear Boolean function then f is a dictatorship. Proof: f(  ),f(i 0 )  {-1,1}, hence Prop: Let f be a balanced Boolean function s.t. as(f)=1, then f is a dictatorship. Proof:, but f is balanced, (i.e. ), therefore f is also linear.

© S.Safra Proving FKN: almost-linear  close to shallow Theorem: Let f:P([n])   be linear, Theorem: Let f:P([n])   be linear, Let Let let i 0 be the index s.t. is maximal let i 0 be the index s.t. is maximalthen Note: f is linear, hence w.l.o.g., assume i 0 =1, then all we need to show is: We show that in the following claim and lemma. Note: f is linear, hence w.l.o.g., assume i 0 =1, then all we need to show is: We show that in the following claim and lemma.

© S.Safra Corollary Corollary: Let f be linear, and then  a shallow boolean function g s.t. Corollary: Let f be linear, and then  a shallow boolean function g s.t. Proof: let, let g be the boolean function closest to l. Then, this is true, as Proof: let, let g be the boolean function closest to l. Then, this is true, as is small (by theorem), is small (by theorem), and additionallyis small, since and additionallyis small, since

© S.Safra Claim 1 Claim 1: Let f be linear. w.l.o.g., assume then  global constant c=min{p,1-p} s.t. Claim 1: Let f be linear. w.l.o.g., assume then  global constant c=min{p,1-p} s.t. {} {1} {2} {i}{n} {1,2} {1,3}{n-1,n}S{1,..,n} weight Characters Each of weight no more than c 

© S.Safra Proof of Claim1 Proof: assume Proof: assume for any z  {3,…,n}, consider x 00 =z, x 10 =z  {1}, x 01 =z  {2}, x 11 =z  {1,2} for any z  {3,…,n}, consider x 00 =z, x 10 =z  {1}, x 01 =z  {2}, x 11 =z  {1,2} then then Next value must be far from {-1,1} ! Next value must be far from {-1,1} ! A contradiction! (to ) A contradiction! (to ) 1 ?

© S.Safra Where to go for Dinner? Who has suggestions: Each cast their vote in an (electronic) envelope, and have the system decided, not necessarily according to majority… It turns out someone –in the Florida wing- has the power to flip some votes Power influence Of course they’ll have to discuss it over dinner….