Fastpoissonslides Using FFT

8/3/2019 Fastpoissonslides Using FFT

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Fast Poisson Solvers and FFT

Tom Lyche

University of Oslo

Norway

Fast Poisson Solvers and FFT p. 1/
http://www.ifi.uio.no/http://www.ifi.uio.no/


2/25

Contents of lecture

Recall the discrete Poisson problem

Recall methods used so far

Exact solution by diagonalization

Exact solution by Fast Sine Transform

The sine transform

The Fourier transform

The Fast Fourier Transform

The Fast Sine Transform

Algorithm



3/25

Recall 5 point-scheme for the Poisson Problem

(u +u ) =f u=0

u=0

u=0

u=0yyxx

h = 1/(m + 1) gives n := m2 interior grid points

Discrete approximation V= [vj,k] Rm,m with vj,k u(jh,kh).Matrix equation TV+ V T = h2F with T = tridiag(

1, 2,

1)

Rm,m

and F = [f(jh,kh)] Rm,m

Linear system Ax = b with A = T V+ V T, x = vec(V) Rnand b = h2vec(F) Rn



4/25

Compare #flops for different methods

System Ax = b of order n = m2 and bandwidth m = n.full Cholesky: O(n3)

band Cholesky, block tridiagonal: O(n2)

If m = 103 then the computing time is

full Cholesky: years

band Cholesky and block tridiagonal , : hours

Derive a method which takes : seconds



5/25

New exact fast method

1. Diagonalization of T

2. Fast Sine transform

restricted to rectangular domains

can be extended to 9 point scheme and biharmonic problem

can be extended to 3D



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Find Vusing diagonalization

We define a matrix X by V= SXS, where V is thesolution of TV+ V T = h2F

TV+ V T = h2F

V=SXS TSXS+ SXST= h2FS( )S

STSXS2 + S2XSTS= h2SFS

TS=SD S2DXS2 + S2XS2D = h2SFSS2=I/(2h)

DX+XD = 4h4SFS.



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X is easy to find

An equation of the form DX+ XD = B for some B iseasy to solve.

Since D = diag(j) we obtain for each entry

jxjk + xjkk = bjk

so xjk = bjk/(j + k) for all j, k.



10/25

Algorithm

Algorithm 1 (A Simple Fast Poisson Solver).

1. h = 1/(m + 1);F =

f(jh,kh)

m

j,k=1;

S=

sin(jkh)mj,k=1; =

sin

2

((jh)/2)mj=1

2. G = (gj,k) = SFS;

3. X= (xj,k)mj,k=1, wherexj,k = h

4gj,k/(j + k);

4. V= SXS;

(1)

Output is the exact solution of the discrete Poisson equation on a

square computed in O(n3/2) operations.

Only a couple of mm matrices are required for storage.Next: Use FFT to reduce the complexity to O(n log2 n)



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The product B = AS

The product B = AScan also be interpreted as a DST.

Indeed, since S is symmetric we have B = (SAT)T

which means that B is the transpose of the DST of the

rows of A.It follows that we can compute the unknowns V inAlgorithm 1 by carrying out Discrete Sine Transforms on

4 m-by-m matrices in addition to the computation of X.



13/25

The Euler formula

ei = cos + i sin , R, i =1

ei = cos

i sin

cos =ei + ei

2, sin =

ei ei2i

Consider

N := e2i/N = cos

2

N i sin

2

N, NN = 1

Note that

jk2m+2 = e2jki/(2m+2) = ejkhi = cos(jkh) i sin(jkh).



14/25

Discrete Fourier Transform (DFT)

If

zj =N

k=1

(j1)(k1)N yk, j = 1, . . . , N

then z = [z1, . . . , zN]T is the DFT of y = [y1, . . . , yN]

T.

z = FNy, where FN := (j1)(k1)N Nj,k=1, RN,N

FN is called the Fourier Matrix



15/25

Example

4 = exp2i/4 = cos(2 ) i sin(2 ) = i

F4 =

1 1 1 1

1 4 24

34

1 24 44

64

1 34 64

94

=

1 1 1 1

1 i 1 i1 1 1 11 i 1 i



16/25

Connection DST and DFT

DST of order m can be computed from DFT of orderN = 2m + 2 as follows:

Lemma 1. Given a positive integerm and a vectorx Rm.Componentk ofS

mx is equal toi/2 times componentk + 1 of

F2m+2z where

z = (0, x1, . . . , xm, 0,xm,xm1, . . . ,x1)T R2m+2.

In symbols

(Smx)k =i

2

(F2m+2z)k+1 , k = 1, . . . , m .



17/25

Fast Fourier Transform (FFT)

Suppose N is even. Express FN in terms of FN/2. Reorder the columnsin FN so that the odd columns appear before the even ones.

PN := (e1, e3, . . . , eN1, e2, e4, . . . , eN)

F4 =

1 1 1 1

1 i 1 i1 1 1 1

1 i 1 i

F4P4 =

1 1 1 1

1 1 i i1 1 1 1

1 1 i i

.

F2 = 1 1

1 2 =

1 1

1 1 , D2 = diag(1, 4) =

1 0

0 i .

F4P4 =

F2 D2F2

F2D2F2



18/25

F2m, P2m, Fm and Dm

Theorem 2. IfN = 2m is even then

F2mP2m =

Fm DmFm

FmDmFm

, (2)

where

Dm = diag(1, N, 2N, . . . ,

m1N ). (3)



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Proof

Fix integers j, k with 0 j,k m 1 and set p = j + 1 and q = k + 1.Since mm = 1,

2N = m, and

mN = 1 we find by considering elements in

the four sub-blocks in turn

(F2mP2m)p,q = j(2k)N = jkm = (Fm)p,q,

(F2mP2m)p+m,q = (j+m)(2k)N =

(j+m)km = (Fm)p,q,

(F2mP2m)p,q+m = j(2k+1)N =

jN

jkm = (DmFm)p,q,

(F2mP2m)p+m,q+m = (j+m)(2k+1)

N = j(2k+1)

N = (DmFm)p,q

It follows that the four m-by-m blocks of F2mP2m have the required

structure.



20/25

The basic step

Using Theorem 2 we can carry out the DFT as a block multiplication.

Let y R2m and set w = PT2my = (wT1 ,wT2 )T, where w1,w2 Rm.Then F2my = F2mP2mP

T2my = F2mP2mw and

F2mP2mw =

Fm DmFmFm DmFm

w1w2

= q1 + q2q1 q2

, whereq1 = Fmw1 and q2 = Dm(Fmw2).

In order to compute F2my we need to compute Fmw1 and Fmw2.

Note that wT1 = [y1, y3, . . . , yN1], while wT2 = [y2, y4, . . . , yN].

This follows since wT = [wT1 ,wT2 ] = y

TP2m and post multiplying a

vector by P2m moves odd indexed components to the left of all theeven indexed components.



21/25

Recursive FFT

Recursive Matlab function when N = 2k

Algorithm 3. (Recursive FFT )

function z=fftrec(y)

n=length(y);

if n==1 z=y;

else

q1=fftrec(y(1:2:n-1));

q2=exp(-2*pi*i/n). (0:n/2-1).*fftrec(y(2:2:n));z=[q1+q2 q1-q2];

end

Such a recursive version of FFT is useful for testing purposes, but is much

too slow for large problems. A challenge for FFT code writers is to develop

nonrecursive versions and also to handle efficiently the case where N is

not a power of two.


FFT C l it


22/25

FFT Complexity

Let xk be the complexity (the number of flops) when N = 2k.

Since we need two FFTs of order N/2 = 2k1 and a multiplication

with the diagonal matrix DN/2 it is reasonable to assume that

xk

= 2xk

1+ 2k for some constant independent of k.

Since x0 = 0 we obtain by induction on k that

xk = k2k = Nlog2 N.

This also holds when N is not a power of 2.

Reasonable implementations of FFT typically have 5


I t i FFT


23/25

Improvement using FFT

The efficiency improvement using the FFT to compute the DFT isimpressive for large N.

The direct multiplication FNy requires O(8n2) flops since complex

arithmetic is involved.

Assuming that the FFT uses 5Nlog2 N flops we find for

N = 220 106 the ratio

8N2

5Nlog2 N 84000.

Thus if the FFT takes one second of computing time and if the

computing time is proportional to the number of flops then the direct

multiplication would take something like 84000 seconds or 23 hours.


P i l b d FFT


24/25

Poisson solver based on FFT

requires O(n log2 n) flops, where n = m2

is the size of the linearsystem Ax = b. Why?

4m sine transforms: G1 = SF, G = G1S, V1 = SX, V= V1S.

A total of 4m FFTs of order 2m + 2 are needed.Since one FFT requires O((2m + 2) log2(2m + 2) flops the 4m FFTs

amounts to 8m(m + 1) log2(2m + 2) 8m2 log2 m = 4n log2 n,

This should be compared to the O(8n3/2

) flops needed for 4straightforward matrix multiplications with S.

What is faster will depend on the programming of the FFT and the

size of the problem.


C t th d


25/25

Compare exact methods

System Ax = b of order n = m2

.Assume m = 1000 so that n = 106.

Method # flops Storage T = 109flops

Full Cholesky 13n3 = 1310

18 n2 = 1012 10 years

Band Cholesky 2n2 = 2 1012 n3/2 = 109 1/2 hourBlock Tridiagonal 2n2 = 2 1012 n3/2 = 109 1/2 hourDiagonalization 8n3/2 = 8 109 n = 106 8 secFFT 20n log2 n = 4 108 n = 106 1/2 sec


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Fastpoissonslides Using FFT