The two-modular Fourier transform of binary functions

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1 of binary functions Yi Hong (Monash University, Melbourne, Australia) (Monash University, Melbourne, Australia) Jean-Claude Belfiore (Telecom ParisTech)

2 The Discrete Fourier Transform N samples of a discrete-time signal (real or complex) form the time-domain vector x = (x[n]) N 1 n=0 The discrete Fourier transform (DFT) of x is the frequency-domain vector X = (X[k]) N 1 k=0 X[k] = N 1 n=0 x[n]e j2πnk N k = 0,...N 1 The inverse discrete Fourier transform (IDFT) of X is x[n] = 1 N N 1 k=0 X[k]e j2πnk N n = 0,...N 1

3 The Discrete Fourier Transform The vector x = (x[n]) N 1 n=0 gives the N samples of a time-domain function f : Z C If f is periodic by N samples then f : Z N C (assumption for DFT to provide the discrete spectrum) The time axis Z N = {0,1,...N 1} is an additive group G = (Z N,+) with addition mod N then f : G C In the frequency domain X = (X[k]) N 1 k=0 represents the transform of f as a periodic function ˆf : G C The frequency axis has the same additive group structure G = (Z N,+)

4 The Discrete Fourier Transform The DFT matrix F = {e j2πnk N } N 1 n,k=0 is a unitary matrix such that X T = Fx T x T = 1 N FH X T The vectors x and X are a two representations of the signal x[n] in different coordinate systems, defined by the time basis and frequency basis. x x[n] X frequency time

5 One-dimensional group representation The Abelian group G = (Z N,+) with addition mod N admits the following one-dimensional representations χ k : G S k C χ k (n) = e j2πnk N where k = 0,...,N 1 and } S k = {1,e j2π k N,e j2π 2k N,,e j2π (N 1)k N The representation χ k is a group homomorphism transforming the addition mod N in G into the multiplication of N-th roots of unity in S k, i.e., for any a,b G χ k (a+b) = χ k (a)χ k (b) since e j2π(a+b)k N = e j2πak N e j2π bk N

6 Example Z 6 k S k ={χ k (g),g G={0,1,2,3,4,5}} Ker(χ k ), G/Ker(χ k ) 0 0 {1} {0,1,2,3,4,5},{0} {1,e j2π 6,e j4π 6,e j6π 6,e j8π 6,e j10π 6 } {0},{0,1,2,3,4,5} {1,e j4π 6,e j8π 6 } {0,3},{0,2,4} {1, 1} {0,2,4},{0,3} {1,e j4π 6,e j8π 6 } {0,3},{0,2,4} {1,e j2π 6,e j4π 6,e j6π 6,e j8π 6,e j10π 6 } {0},{0,1,2,3,4,5} 4 5

7 Example Z 6 (cont.) g G χ 0 (g) ψ 0 χ 1 (g) 1 e j2π 6 e j4π 6 e j6π 6 e j8π 6 e j10π 6 ψ 1 χ 2 (g) 1 e j4π 6 e j8π 6 1 e j4π 6 e j8π 6 ψ 2 χ 3 (g) ψ 3 χ 4 (g) 1 e j4π 6 e j8π 6 1 e j4π 6 e j8π 6 ψ 4 χ 5 (g) 1 e j2π 6 e j4π 6 e j6π 6 e j8π 6 e j10π 6 ψ 5 F = ψ 0. ψ 5

8 DFT using representations as Fourier basis The representations χ k for k = 0,...,N 1 are all inequivalent. Some are one-to-one and some are many-to-one and the images can be associated with subgroups of G We can formally rewrite the DFT as X[k] = x,ψ k = g Gx[g]χ k (g) k = 0,...,N 1 The complex vectors ψ k = [χ k (g)] g G form the discrete Fourier basis vectors Each representation provides a lens through which we observe the time-domain signal x[n].

9 FFT using representations as Fourier basis Given a normal subgroup H G we define the quotient group G/H consisting of the coset leaders u of the cosets u+h The direct product of H and G/H is isomorphic to G i.e., G = {u+v u H,v G/H} H G/H All u H = Ker(χ k ) are mapped to the same value χ k (u) = χ k (0) = 1 S k Then we can compute the DFT more efficiently as X[k] = x[g]χ k (g) = x[u+v]χ k (u+v) g G v G/H u H = ( ) x[u+v] χ k (v) k = 0,...,N 1 v G/H u H

10 Known generalizations of the DFT concept f : G C, where G can be an arbitrary group (not only Abelian): Fourier coefficients are complex matrices These generalizations make use of multi-dimensional representations of the group G with matrices over C The inverse Fourier transform uses 1 G Tr( ) the Trace operator of a matrix to get back to time domain scalar values of f. f : G K, where K is a field of characteristic p and p does not divide G : Fourier coefficients are scalars in K since an exponential function can be defined using a primitive element α K.

11 The missing Fourier Transform for binary functions We consider a finite commutative ring R of characteristic p = 2, e.g., F 2 [X]/φ(X), where φ(x) is a binary-coefficient polynomial of degree m. Elements of R are represented by m-bit vectors (or polynomials of degree at most m 1) and multiplications are computed by polynomial multiplication mod φ(x). Let G = C2 n be the additive group of Fn 2 (n bit vectors) We study binary functions f : G R (n bit to m bit) and their convolutions (group ring R[G]) What does not work? If p = 2 divides G = 2 n = N, the inverse DFT term 1/N is not defined in R and the Trace fails to work in the generalized inverse DFT.

12 The two-modular representations of G Let G = C 2 = {0,1} then a two-modular representation as 2 2 binary matrices over R, is given by the two matrices ( ) ( ) E 0 = π 1 (0) = and E = π 1 (1) = 0 1 The matrix entries are the zero and one element in R. The n-fold direct product of C 2, G = C n 2 = C 2 C 2 can be represented as the Kroneker product of the representations of C 2, i.e., π n (G) π 1 (C 2 ) π 1 (C 2 )

13 The two-modular representations of G = C 2 C 2 Let the binary vectors b = (b 1,...,b n ) represent the elements of G with bitwise addition mod 2 (XOR). Then π n (b) = E b = π 1 (b 1 ) π 1 (b n ) Example G = C 2 C 2 ( ) ( ) E 00 = E 01 = E 10 = ( ) E 11 = ( )

14 The two-modular Fourier basis for G = C 2 C 2 Consider the two-modular representations (2 k 2 k matrices) of the nested subgroups of G = C n 2, H 0 = {0 n } H 1 H k H n 1 H n = G where H 1 = C2, H 2 = C2 C 2, H 3 = C2 C 2 C 2, etc. The Fourier basis vectors are made up of all the inequivalent two-modular representations π k H 0 = Im(π0 ) = {1} H 1 = Im(π1 ) = {E 0,E 1 } H 2 = Im(π2 ) = {E 00,E 01,E 10,E 11 } H 3 = Im(π3 ) = {E 000,E 001,E 010,E 011,E 100,E 101,E 110,E 111 }.

15 Example G = C 3 2 The Fourier basis vectors ψ k = [E τk (g) : g G] are the 2 n -component vectors (indexed by g) of 2 k 2 k matrices from the set Im(π k ) g G π 0 (g) ψ 0 π 1 (g) E 0 E 1 E 0 E 1 E 0 E 1 E 0 E 1 ψ 1 π 2 (g) E 00 E 01 E 11 E 10 E 00 E 01 E 11 E 10 ψ 2 π 3 (g) E 000 E 010 E 001 E 011 E 111 E 101 E 110 E 100 ψ 3

16 The two-modular Fourier Transform of f : G R We abstractly define the k-th Fourier coefficients as the 2 k 2 k matrix with elements in R ˆf k = f,ψ k g Gf(g)E τk (g) for k = 0,...,n (1) where E τk (g) ψ k selected by g according to a (surjective) group homomorphism τ k : G C2 k (n-bit to k-bits).

17 The two-modular Fast Fourier Transform We can represent the elements of H k = C k 2 as n-bit vectors with the first n k bits set to zero i.e., H k = {(0,...,0,b n k+1,...,b n ) b i {0,1}} = C k 2 k = 1,...,n The quotient groups G/H k are represented as n-bit vectors with the last k bits set to zero i.e., { {(b1,...,b G/H k = n k,0,...,0) b i {0,1}} k = 1,...,n 1 {0 n } k = n. Consider the binary subgroup d k = {0,d k } of H k where Then d k = (0,...,0,b n k+1 = 1,0...,0) }{{} G H k / d k }{{} 2 n d k }{{} 2 k 1 2 G/H k }{{} 2 n k

18 The two-modular Fast Fourier Transform The k-th Fourier coefficients ˆf k can be explicitly computed by collecting the terms with the same E τk (g), i.e., ˆf k = f(u+v) v G n/h k + f(u+d k +v) v G n/h k u H k / d k E σk (u) E σk (u) k = 0,...,n where σ k (u) is a map converting the n bit vector u H k / d k to a k bit vector in C2 k, i.e., it removes the first n k zero bits of the n bit vector u. The corresponding σ k (u) is the binary complement of the bits of σ k (u).

19 The two-modular Inverse Fourier Transform Let π k (g) = E τk (g) be the 2 k 2 k representation of an element g C2 k then we define the character of g as Φ k (E τk (g)) (E τk (g)) (1,2 k ) {0,1} i.e., Φ extracts the top-right corner element of the matrix E g. The representation of the all ones vector 1 = (1,1,...,1) yields Φ(E 1 ) = 1, while any other binary vector representation is mapped to zero. Since Φ k is an homomorphism, we have { 1 iff a b = 1 (or a = b) Φ k (E a E b ) = Φ k (E a b ) = 0 otherwise

20 The two-modular Inverse Fourier Transform The inverse Fourier transform is given by f j = f c = ˆf n ) 0 + Φ k (ˆfk E τk (c) k=1 j = 0,...,2 n 1 (2) where c = (c n,...,c k,...,c 1 ) and j = D(c) is the decimal representation of c.

21 The convolution theorem can be used to transform convolution in the time-domain defined as (f 1 f 2 )(a) b G f 1 (ab 1 )f 2 (b) = b Gf 1 (a b)f 2 (b) into the product in the frequency domain i.e., (f 1 f 2 )(ρ) = ˆf 1 (ρ)ˆf 2 (ρ) This is where the multiplicative structure of R plays a role.

22 Conclusions A new look at the Discrete Fourier Transform Potential applications of the two-modular Fourier transform reliable computation of binary functions classification of binary functions (polynomial vs. transcendental over R) cryptography complexity of binary functions

23 Thank you!

The Two-Modular Fourier Transform of Binary Functions

1 The Two-Modular Fourier Transform of Binary Functions Yi Hong, Senior Member, IEEE, Emanuele Viterbo, Fellow, IEEE, and Jean-Claude Belfiore, Member, IEEE Abstract In this paper, we provide a solution