Fourier Analysis,
Projections,
Influences,
Juntas,
Etc…
Boolean Functions and Juntas
A boolean function
f : P n 1,1
Def: f is a j-Junta if there exists J[n]
where |J|≤ j, and s.t. for every x
f(x) = f(x J)
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f is (, j)-Junta if j-Junta f’ s.t. Pr f x f' x
x
Functions as an
Inner-Product Vector-Space
11*
1*
10*
*
01*
0*
00*
111*
110*
101*
100*
011*
010*
001*
000*
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f
2n
11*
11*
Functions as an
Inner-Product Vector-Space
A functions f is a vector
2n
f
Inner product (normalized)
fg
x2n
Norm (normalized)
fp
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E f x g x
p 1/p
fx
E
n
x2
1*
1*
10*
10*
**
01*
01*
0*
0*
00*
00*
111*
111*
110*
110*
101*
101*
100*
100*
011*
011*
010*
010*
001*
001*
000*
000*
ff
11*
11*
1*
1*
10*
10*
Simple Observations
Claims:
f 1 E f(x)
x
For a boolean f
p
fp 1
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**
01*
01*
0*
0*
00*
00*
111*
111*
110*
110*
101*
101*
100*
100*
011*
011*
010*
010*
001*
001*
000*
000*
ff
11*
11*
1*
1*
10*
10*
Fourier-Walsh Transform
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00*
00*
Sx
Given any function f : P n
let the Fourier-Walsh coefficients of f be
f S f S
01*
01*
0*
0*
Consider all multiplicative functions, one for
each character S[n]
S (x) 1
**
thus f can described as
f f SS
S
111*
111*
110*
110*
101*
101*
100*
100*
011*
011*
010*
010*
001*
001*
000*
000*
ff
11*
11*
1*
1*
10*
10*
Fourier Transform: Norm
Norm: (not normalized)
f
Thm [Parseval]:
p
p
f S
01*
01*
0*
0*
p
Sn
f f2
Hence, for a boolean f
2
2
f (S)
S
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**
2
f2 1
00*
00*
111*
111*
110*
110*
101*
101*
100*
100*
011*
011*
010*
010*
001*
001*
000*
000*
ff
11*
11*
1*
1*
10*
10*
Simple Observations
Claim:
Hence, for any f
00*
00*
011*
011*
010*
010*
001*
001*
000*
000*
x
V f E f
2
xP n
xP n
2
f 2 f
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01*
01*
0*
0*
101*
101*
100*
100*
f E f(x)
2
**
111*
111*
110*
110*
2
x E f x
Sn,S
2
f S
ff
Putting a Junta to the Test
Joint work with Eldar Fischer & Guy Kindler
Building on [KKL,Freidgut,Bourgain]
Junta Test
Def: A Junta test is as follows:
A distribution over l queries : P n 0,1
For each l-tuple, a local-test that either accepts or
rejects:
T[x1, …, xl]: {1, -1}l{T,F}
l
s.t. for a j-junta f
Prx1 ,..,xl T x1 ,.., xl f 1
whereas for any f which is not (, j)-Junta
Prx1 ,..,xl T x1 ,.., xl (f) 1 2
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Variables` Influence
The influence of an index i [n] on a boolean
function f:{1,-1}n {1,-1} is
Influencei (f) Pr f x f x i
xPn
Which can be expressed in terms of the
Fourier coefficients of f
Claim:
2
Influence i f f S
iS
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Fourier Representation of influence
Proof: consider the I-average function on P[I]
AI f (x) E f x y
yPI
which in Fourier representation is
and
AI f
SI
f(S) S
2
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2
influencei f 1 Ai f f (S)
2
iS
High vs Low Frequencies
Def: The section of a function f above k is
f
k
f S
S k
S
and the low-frequency portion is
f
k
f S
S k
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S
Subsets` Influence
Def: The influence of a subset I [n] on a
boolean function f is
InfluenceI f 1 AI f
2
2
2
f S
SI
and the low-frequency influence
Influence
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k
I
f InfluenceI f
k
SI
S k
2
f S
Independence-Test
The I-independence-test on a boolean function
w I, z1 ,z2 I
f is, for
IT(w, z1 ,z2 ) :
?
f w z1 f w z2
Lemma:
Pr IT(w, z1 ,z2 ) 1 21 influenceI f
wP I
z1 ,z2 PI
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Pr IT(w, z1 ,z2 ) 1 21 influenceI f
wP I
z1 ,z2 PI
Pr IT(x, y1 ,y
,y2 ) E
xP[n]
P[n]
xP I
y1 ,y2 PI
1AI f x
2
2
1 AI f x
2
2
1
E
1
1
A
f
I
2
2
xP[n]
1 21 influence I f
2
22 AI f x
4
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2
Junta Test
The
junta-size test on a
boolean function f is
Randomly
partition [n] to I1, .., Ir
Run the independence-test t
times on each Ih
Accept if ≤j of the Ih fail their
independence-tests
For r>>j2 and t>>j2/
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Completeness
Lemma: for a j-junta f
Pr IT(x, y1,y2 ) 1
xP I
y1 ,y2 PI
Proof: only those sets which contain
an index of the Junta would fail
the independence-test
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Soundness
Lemma:
Pr IT(x, y1 ,y2 ) 1
2
xP I
y1 ,y2 PI
f is an
( , j) junta
Proof: Assume the premise. Fix <<1/t and let
J i| influencei f
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|J| ≤ j
Prop: r >> j implies |J| ≤ j
Proof: otherwise,
J spreads among Ih w.h.p.
and for any Ih s.t. IhJ ≠ it must be that
influenceI(f) >
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High Frequencies Contribute Little
Prop: k >> r log r implies
f
k 2
2
2
f
S
S k
4
Proof: a character S of size larger than k
spreads w.h.p. over all parts Ih, hence
contributes to the influence of all parts.
If such characters were heavy (>/4), then
surely there would be more than j parts Ih
that fail the t independence-tests
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Almost all Weight is on J
Lemma: influencek f
J
4
Proof: otherwise,
k
k
since
influencei f influenceJ f
iJ
for a random partition w.h.p. (Chernoff bound)
k
for every h
influence f
iIh
i
100r
however, since for any I
influence f k influence f
iI
k
i
k
I
the influence of every Ih would be ≥ /100rk
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Find the Close Junta
Now, since
influenceJ f influence
k
J
f
f
k 2
2
consider the (non boolean)
g
f S
S J
S
which, if rounded outside J
f' x sign AJ f x J
is boolean and not more than far from f
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2
Open Problems
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Is there a characterization, via Fourier
transform, of all efficiently testable
properties?
What about tests that probe f only at
two or three points? With applications
to hardness of approximation.
Product, Biased Distribution
Consider the q-biased product
distribution q:
Def: The probability of a subset F
F q (1 q)
F
n
q
n F
and for a family of subsets
qn Pr F qn F
n
Fq
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F
Beckner/Nelson/Bonami
Inequality
Def: let T be the following operator on any f,
T f x
Prop:
Proof:
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T f
T f x
E
z 1 / 2
f x z
f S
S
Sn
f S
Sn
S
S
x E S z
z
Beckner/Nelson/Bonami
Inequality
Def: let T be the following operator on any f,
T f x
E
z 1 / 2
f x z
Thm: for any p≥r and ≤((r-1)/(p-1))½
T f f r
p
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Beckner/Nelson/Bonami Corollary
Corollary: for f s.t. f>k=0 and p≥r≥1
k
f
p
p 1
r 1
2
fr
Proof:
fp
k
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f S
S
Sn
S
p
T f p f r
Average Sensitivity
The sum of variables’ influence is referred to
as the average sensitivity
as f
influence f
i
i[n]
Which can be expressed by the Fourier
coefficients as
2
as f f (S) S
S
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Freidgut Theorem
Thm: any boolean f is an [, j]-junta for
O as f /
j =2
Proof:
1.
2.
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Specify the junta J
Show the complement of J has little influence
Specify the Junta
Set k=O(as(f)/), and =2-O(k)
Let J i| influencei f
We’ll prove:
and let
2
avgJ f 1
2
2
f'(x) sign avgJ f x J
hence, J is a [,j]-junta, and |J|=2O(k)
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High Frequencies Contribute Little
Prop:
f
k 2
2
2
f
S
S k
4
Proof: a character S of size larger than
k contributes k times the square of its
coefficient to the average sensitivity.
If such characters were heavy (>/4),
as(f) would have been large
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Altogether
Lemma: influenceJ f
2
Proof:
influenceJ f f
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k 2
2
+ influence Jk f
2
Altogether
k
k
k
k
influence
f
influence
influ
ence
influence JJ iinf
luence ii f
nflu
ii
J
J
ii
S,
S, S
S
k
k
J
J ii
2
O(k)
O(k)
2
O(k)
O(k)
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ii
J
J
2
2
f(S) SS
2
2
f(S
f(S))
f(S)
ii
S
S
O(k)
2O(k)
ii
S,
S, S
S
k
k
J
J ii
2
2
S
S
2
O(k)
O(k)
rr
influence
in
influenc
influ
encee ii f
inffluence
luenc
2/
2/ rr
ii
J
ii
J
J
2
O(k)
O (k)
O(k)
2
2
f(S) SS
f(S)
f(S)
ii
S
S
rr
4
4 // rr
S
S
2
2
as f
2r
r
Biased q-Influence
The q-influence of an index i [n] on a boolean
function f:P[n] {1,-1} is
Influenceiq (f) Prn f x f x
i
xq
influenceiq f
asq f
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1
n
q
Ai f
q
influence
i f
i1
2
2
Thm [Margulis-Russo]:
For monotone
asq ( )
d q ( )
dq
Hence
Lemma:
For monotone
> 0, q[p, p+] s.t. asq() 1/
Proof: Otherwise p+() > 1
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Proof [Margulis-Russo]:
dq ( )
dq
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n
i1
qi ( )
qi
n
influence asq ( )
i1
q
i
Erdös-Ko-Rado
Def: A family of subsets
P[R] is t-intersecting if for every
F1, F2 , |F1 F2| t
p (P) = max p (Ai,2 )
i
Thm[Wilson,Frankl,Ahlswede-Khachatrian]:
For a t-intersecting ,
where
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p ( ) max p (Ai,t )
i
Ai,t F | F 1,...,2i t i t
Corollary: p() > P is not 2-intersecting
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