Analysis of Boolean Functions Fourier Analysis, Projections, Influence, Junta, Etc… Slides prepared with help of Ricky Rosen

Introduction Objectives: Objectives: Codes and Juntas using Fourier Analysis. Codes and Juntas using Fourier Analysis. Overview: Overview: Codes basic definitions Codes basic definitions Testing Hadamard code Testing Hadamard code Testing Long code Testing Long code Junta Test Junta Test

Codes and Boolean Functions Def: an m-bit binary code is a subset of the set of all m-binary strings C  {-1,1} m The distance of a code C, 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

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

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

6 Hadamard Test – Soundness Prop(soundness): Proof: if f(x)  f(y)=f(xy), then f(x)  f(y)  f(xy)=1 else f(x)  f(y)  f(xy)=-1  x,y [f(x)f(y)f(xy)]= 1  Pr[f(x)f(y)=f(xy)] -1  Pr[f(x)f(y)  f(xy)]  ½(1+  ) -½(1-  )= 

proof

Proof cont. Conclusion: Conclusion: Proof 1: (probabilistic method) – consider random variables that with probability are valued And its expectation is >  then one of its variables > . Proof 1: (probabilistic method) – consider random variables that with probability are valued And its expectation is >  then one of its variables > . Proof 2: (algebraic): if then Proof 2: (algebraic): if then How large can be? How large can be?

Juntas A function is a J-junta if its value depends on only J variables. A function is a J-junta if its value depends on only J variables. A Dictatorship is 1-junta A Dictatorship is 1-junta

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

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

Testing Long-code Def(a long-code list-test): given a code-word f, probe it in a constant number of entries, and Completeness (not perfect): accept almost always if f is a monotone dictatorship Completeness (not perfect): accept almost always if f is a monotone dictatorship Soundness: 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. Soundness: 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.

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.

What about a Hadamar like test? completeness? completeness?yes Soundness? Soundness? We would like something like: Which functions will pass the test? Which functions will pass the test? all the characters for start all the characters for start and many more… and many more…no

Perturbation Def: denote by  p 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-p i  x with probability 1-p i  x with probability p i  x with probability p

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- 

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

Proof cont. Try to find k s.t. Try to find k s.t. to get k can be determined according to  and the size of the character.

List decoding This test does not allow list-decoding a function. Problem: the function   passes the test, as well as functions close to it. Solution (?) assume f is folded (odd): f(x)=-f(-x) for every x. (make sure you understand why this is a “solution”)

Junta Test (1) Definitions (2) Independence test (3) The size test (4) Soundness and completeness of the tests

Definitions: Variation Def: the variation of f (extension of influence ) Intuition: if I is very influential on f then the function will go “wild” on y  P[I] hence the expected variance (  variation) is large.

Variation cont. Prop: the following is an equivalent definitions to the variation of f: Recall

Recall the variance of f Recall the variance of f Hence Hence

Proof – Cont. Recall Recall Therefore (by Parseval): Therefore (by Parseval):

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

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

Fourier Representation of influence Recall: consider the I-average function on P[I] which in Fourier representation is and

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

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

proof

What was I looking for?

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 for r>>j 2 Randomly partition [n] to I 1,.., I r for r>>j 2 Run IT t times on each I h for t>>j 2 /  Run IT t times on each I h for t>>j 2 /  Accept if no more than j of the I h fail IT Accept if no more than j of the I h fail IT

Completeness completeness: for a j-junta f only those I h that contain a member of the Junta fail IT.  No more than j sets can fail the test.

Soundness Soundness: if f passes the test w.p. ½ then f is ( ,j) -junta Proof: utilize bounds on the variation of those I h that pass IT. Intuition: A set, I h, has probability of ½Variation to fail IT once. If I h passes IT t times, one expects that ½Variation( I h ) < 1/t

Formally (if you insist): The probability of the event that I h fails IT is p= ½Variation. Formally (if you insist): The probability of the event that I h fails IT is p= ½Variation. The probability of I h to pass IT t times is (1-p) t. The probability of I h to pass IT t times is (1-p) t. If it happens w.h.p then If it happens w.h.p then e -pt > (1-p) t > ½ -pt > ln(½) p < 1/t(-ln(½)) < 1/t

Soundness Proof Assume the premise. Fix  >1/t and let using the bound on p we prove that if f passes JT then f is  close to a J–junta. Prop: if JT succeeds w.p > ½ then |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) > 

j spread For a random partition, by birthday problem, for r>j 2 and fix some j variables from J, w.h.p. no two members of J fall in the same I h. For a random partition, by birthday problem, for r>j 2 and fix some j variables from J, w.h.p. no two members of J fall in the same I h. Choose r s.t. w.p.  ¾ at least j+1 members of J are spread in distinct I h ’s. Choose r s.t. w.p.  ¾ at least j+1 members of J are spread in distinct I h ’s.

 I containing one of the variables in J and a fixed i  I: and since I contains a variable of J, variation I >  Since j 2 /  2/t[ln(j+1)+ln4] Since j 2 /  2/t[ln(j+1)+ln4] Now, for a random partition one (like you) can bound the probability that one of the I h that contain of of the j+1 members of J passes IT t times by the union bound: Now, for a random partition one (like you) can bound the probability that one of the I h that contain of of the j+1 members of J passes IT t times by the union bound: The probability of the size test to succeed is < ¼ + ¾  ¼ < ½ The probability of the size test to succeed is < ¼ + ¾  ¼ < ½ contradiction to the assumption that the test succeeds w.p >½ J does not spread between j+1 I’s J does spreads between j+1 I’s and IT succeeds Contradictio n to what??

Where are we? We concluded that if the JT succeeds w.p > ½ then |J| ½ then |J|<j Now what? Now what? We will show that almost all the weight of f is concentrated on J. How ? How ? (1) Show that the total weight on the high frequencies is small. (2) Show that the total weight of the low frequencies on the characters that are not contained in J is small.

(1)High Frequencies Contribute Little Prop: k >> r log r implies Proof: by the Coupon Collector Problem, a character S of size larger than k spreads w.h.p. (>¾) over all the I h (namely, intersects every I h ), hence contributes to the influence of all parts. For this event: In every I h  member of S (S s.t. |S|>k)

High frequencies cont. Use union bound to bound the probability that one of I 1 … I j+1 to pass IT t times test. This probability is <  w.p. at least ¾  ¾ = 9/16 JT fails. contradiction contradiction J 2 >ln(j+1)+ln4 8j 2 /  <t Prob. for spreading over all I h Prob. That the first j+1 groups fail the size test

(2)Almost all Weight is on J Lemma: Proof: assume by way of contradiction otherwise since for a random partition w.h.p. (>¾) ( by a Chernoff like bound – (  i  influence i ¾) ( by a Chernoff like bound – (  i  influence i <  ) for every h however, since for any I And also

Similar to the last claim, the probability to fail the test in such an event is at least ¾. Similar to the last claim, the probability to fail the test in such an event is at least ¾.  the test fails w.p > ½  the test fails w.p > ½contradiction Note: for this union bound t=200rk/  [ln(j+1)+ln4] Note: for this union bound t=200rk/  [ln(j+1)+ln4]

Find the Close Junta Now, since consider the (non Boolean) which, if rounded outside J

Then Then The distance of f’ from g --the closest Boolean function to g-- is no more than f’s The distance of f’ from g --the closest Boolean function to g-- is no more than f’s By the triangle inequality By the triangle inequality

Juntas A function is a J-junta if its value depends on only J variables. A function is a J-junta if its value depends on only J variables. A Dictatorship is 1-junta A Dictatorship is 1-junta

- Noise sensitivity - Noise sensitivity The noise sensitivity of a function f is the probability that f changes its value when changing a subset of its variables according to the  p distribution. The noise sensitivity of a function f is the probability that f changes its value when changing a subset of its variables according to the  p distribution. Choose a subset (I) of variables Each var is in the set with probability Choose a subset (I) of variables Each var is in the set with probability Flip each value of the subset (I) with probability p Flip each value of the subset (I) with probability p What is the new value of f? I I

Noise sensitivity and juntas  Juntas are noise insensitive (stable) Thm [Bourgain; Kindler & S]: Noise insensitive (stable) Boolean functions are Juntas Choose a subset (I) of variables Each var is in the set with probability Choose a subset (I) of variables Each var is in the set with probability Flip each value of the subset (I) with probability p Flip each value of the subset (I) with probability p What is the new value of f? W.H.P STAY THE SAME What is the new value of f? W.H.P STAY THE SAME I I Junta

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

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:

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

Low-degree B.f are Juntas [KS] 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- (½) 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 )

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

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 is the minimum, over all pairs of legal-words (in C), of the Hamming distance between the two words Note: 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

Long-Code  Monotone-Dictatorship In the long-code, the legal code-words are all monotone dictatorships C={  {i} | i  [n]} namely, all the singleton characters In the long-code, the legal code-words are all monotone dictatorships C={  {i} | i  [n]} namely, all the singleton characters

Open Questions Mechanism Design: show a non truth-revealing protocol in which the pay is smaller (Nash equilibrium when all agents tell the truth?) Mechanism Design: show a non truth-revealing protocol in which the pay is smaller (Nash equilibrium when all agents tell the truth?) Hardness of Approximation: Hardness of Approximation: MAX-CUT MAX-CUT Coloring a 3-colorable graph with fewest colors Coloring a 3-colorable graph with fewest colors Graph Properties: find sharp-thresholds for properties Graph Properties: find sharp-thresholds for properties Analysis: show weakest condition for a function to be a Junta Analysis: show weakest condition for a function to be a Junta Apply Concentration of Measure techniques to other problems in Complexity Theory Apply Concentration of Measure techniques to other problems in Complexity Theory

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)

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)

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

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

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

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.

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.

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

Altogether Lemma: Proof:

Altogether

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

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

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

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 

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- 

Long-code Tests Def (a long-code test): given a code- word w, probe it in a constant number of entries, and Def (a long-code test): given a code- word w, probe it in a constant number of entries, and accept w.h.p if w is a monotone dictatorship accept w.h.p if w is a monotone dictatorship reject w.h.p if w is not close to any monotone dictatorship reject w.h.p if w is not close to any monotone dictatorship

Efficient Long-code Tests For some applications, it suffices if the test may accept illegal code-words, nevertheless, ones which have short list-decoding: Def(a long-code list-test): given a code-word w, probe it in 2/3 places, and accept w.h.p if w is a monotone dictatorship, accept w.h.p if w is a monotone dictatorship, reject w.h.p if w is not even approximately determined by a short list of domain elements, that is, if  a Junta J  [n] s.t. f is close to f’ and f’(x)=f’(x  J) for all x reject w.h.p if w is not even approximately determined by a short list of domain elements, that is, if  a Junta J  [n] s.t. f is close to f’ and f’(x)=f’(x  J) for all x Note: a long-code list-test, distinguishes between the case w is a dictatorship, to the case w is far from a junta.

General Direction These tests may vary These tests may vary The long-code list-test a, in particular the biased case version, seem essential in proving improved hardness results for approximation problems The long-code list-test a, in particular the biased case version, seem essential in proving improved hardness results for approximation problems Other interesting applications Other interesting applications Hence finding simple, weak as possible, sufficient-conditions for a function to be a junta is important. Hence finding simple, weak as possible, sufficient-conditions for a function to be a junta is important.