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Fourier Transforms in Computer Science. Can a function [0,2 p ] z R be expressed as a linear combination of sin nx, cos nx ?. If yes, how do we find the coefficients?. S. If f(x)= a exp(2 p i n x). n. n Z. Fourier’s recipe. then. a = f(x)exp(-2 p i n x) dx. n.
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Fourier Transforms in Computer Science
Can a function [0,2p]zR be expressed as a linear combination of sin nx, cos nx ? If yes, how do we find the coefficients?
S If f(x)= a exp(2pi n x) n n Z Fourier’s recipe then a = f(x)exp(-2pi n x) dx n The reason that this works is that the exp(-2pi nx) are orthonormal with respect to the inner product <f,g> = f(x)g(x) dx
Given “good” f:[0,1]zC we define its Fourier transform as f:ZzC - - f(n) = f(x)exp(-2pi n x) dx space of functions space of functions Fourier Transform
space of functions space of functions Fourier Transform 2 2 L [0,1] L (Z) isometry - - Plancherel formula <f,g> = <f,g> - Parseval’s identity ||f|| = ||f|| 2 2
space of functions space of functions Fourier Transform pointwise multiplication convolution - - f g f*g (f*g)(x)= f(y)g(x-y) dy
Can be studied in a more general setting: LCA group G interval [0,1] Lebesgue measure and integral Haar measure and integral characters of G exp(2pi n x) form a topological group G, dual of G continuous homomorphisms GzC -
Fourier coefficients of AC functions 0 Linial, Mansour, Nisan ‘93
circuit output AND depth=3 OR OR OR AND AND AND AND size=8 x x x x x x Input: 1 2 3 1 2 3
circuits AC 0 constant depth, polynomial size output AND depth=3 OR OR OR AND AND AND AND size=8 x x x x x x Input: 1 2 3 1 2 3
Random restriction of a function n f : {0,1} z {0,1} x x . . . x 1 2 n x 1 0 2 (1-p)/2 (1-p)/2 p
n Fourier transform over Z 2 characters x for each subset of {1,...,n} P i c (x)= (-1) A i A Fourier coefficients - c c f(A)= P(f(x)= (x))-P(f(x)= (x)) A A
Hastad switching lemma o 0 f AC z high Fourier coefficients of a random restriction are zero with high probability All coefficients of size >s are 0 with probability at least s 1/d 1-1/d 1-M(5p s ) depth size of the circuit
We can express the Fourier coef- ficients of the random restriction of f using the Fourier coefficients of f - - |x| E[r(x)]=p f(x) - - S 2 |x| 2 |y| E[r(x) ]=p f(x+y) (1-p) ymx
Sum of the squares of the high Fourier coefficients of an AC Function is small 0 - 1 1/d S 2 f(x) < 2M exp(- (t/2) ) 5e |x|>s 0 Learning of AC functions
Influence of variables on Boolean functions Kahn, Kalai, Linial ‘88
The Influence of variables I (x ) f i The influence of x on f(x ,x , …,x ) i n 1 2 set the other variables randomly the probability that change of x will change the value of the function i Examples: for the AND function of n variables each variable has influence 1/2 for the XOR function of n variables each variable has influence 1 n
The Influence of variables I (x ) f i The influence of x in f(x ,x , …,x ) i n 1 2 set the other variables randomly the probability that change of x will change the value of the function i For balanced f there is a variable with influence > (c log n)/n
We have a function f such that I(x )=||f || , and the Fourier coef- ficients of f can be expressed using the Fourier coefficients of f i p i i p i f (x)=f(x)-f(x+i) i - - f (x) =2f(x) if i is in x = 0 otherwise i
We can express the sum of the influences using the Fourier coefficients of f - S S 2 I(x ) 4 |x| f(x) = i if f has large high Fourier coefficients then we are happy How to inspect small coefficients?
Beckner’s linear operators - - |x| f(x) a f(x) a<1 Norm 1 linear operator from L (Z ) to L (Z ) 2 n 2 n 1+a 2 2 Can get bound ignoring high FC - S 4/3 S |x| 2 I(x ) 4 |x| f(x) (1/2) > i
Explicit Expanders Gaber, Galil ’79 (using Margulis ’73)
Expander Any (not too big) set of vertices W has many neighbors (at least (1+a)|W|) positive constant W
Expander Any (not too big) set of vertices W has many neighbors (at least (1+a)|W|) positive constant N(W) W |N(W)|>(1+a)|W|
Why do we want explicit expanders of small degree? extracting randomness sorting networks
Example of explicit bipartite expander of constant degree: Z x Z Z x Z m m m m (x,y) (x+y,y) (x,y) (x+y+1,y) (x,x+y) (x,x+y+1)
Transform to a continuous problem M(s(A)-A)+M(t(A)-A)>2cM(A) 2 T measure For any measurable A one of the transformations s:(x,y)z(x+y,y) and t:(x,y)z(x,x+y) “displaces” it
Estimating the Rayleigh quotient of an operator on X L(T ) 2 2 m - Functions with f(0)=0 (T f) (x,y)=f(x-y,y)+f(x,y-x) r(T)=sup { |<Tx,x>| ; ||x||=1}
It is easier to analyze the corres- ponding linear operator in Z 2 (S f)(x,y)=f(x+y,y)+f(x,x+y) Let L be a labeling of the arcs of the graph with vertex set Z x Z and edges (x,y)z(x+y,y) and (x,y)z(x,x+y) such that L(u,v)=1/L(v,u). Let C be maximum over all vertices of sum of the labels of the outgoing edges. Then r(S)cC.
Lattice Duality: Banaszczyk’s Transference Theorem Banaszczyk ‘93
Lattice: given n x n regular matrix B, a lattice is { Bx; x Z } n
Successive minima: • smallest r such that a ball centered in 0 of diameter r contains k linearly independent lattice points k
Dual lattice: -T Lattice L with matrix B * Transference theorem: l l c n * k n-k+1 can be used to show that O(n) approximation of the shortest lattice vector in 2-norm is not NP-hard unless NP=co-NP (Lagarias, H.W. Lenstra, Schnorr’ 90)
Poisson summation formula: For “nice” f:RzC - S S f(x) = f(x) x Z x Z
Define Gaussian-like measure on the subsets of R n S 2 r(A)= exp(-p ||x|| ) x A Prove using Poisson summation formula: r((L+u)\B)<0.285 r(L) a ball of diameter (3/4)n centered around 0 1/2
Define Gaussian-like measure on the subsets of R p(A)= r(A L)/r(L) Prove using Poisson summation formula: - p(u) = r(L+u)/r(L) * *
l l > n * If then we have k n-k+1 a vector u perpendicular to all lattice points in L B and L B outside ball * small + small inside ball, “moved by u” r(L +u)/r(L ) * * - S T p(u)= r(x)exp(-2piu x) x L large
Weight of a function – sum of columns (mod m) , Therien ‘94
m 2 4 1 2 4 1 4 1 1 3 4 1 3 4 1 4 n 3 6 1 3 6 1 2 1 3 2 3 2 4 2 5 5 Is there a set of columns which sum to the zero column (mod t) ?
If m>c n t then there always exists a set of columns which sum to zero column (mod t) 11/2 wt(f)= number of non-zero Fourier coefficients w(fg)cw(f)w(g) w(f+g)cw(f)+w(g)
t+1 is a prime x a x a + ... + 1 i,1 m i,m f (x)=g i If no set of columns sums to the zero column mod t then P t g= (1-(1-f ) ) i m restricted to B={0,1} is 1 0 m m n t = wt(g 1 )c t (t-1) B