**
Clarification of Question by
craze4you-ga**
on
20 Nov 2005 17:23 PST
ps
maybe the code below may be helpful, but i do not know the language it
is written it.
---------------------------------
FUNCTION Voss_RandomAdditions, N
; generate Voss fractional Browninan motion in two dimension.
; see"Fractals" by Jen Feder page 180
; R. F. Voss, "Random Fractal Forgeries", in "Fundimental Algorithms for
Computer Graphics",
; editor R. A. Earnshaw, NATO ASI Series F, Computer and System Sciences,
Vol 17, 1985.
; This uses the Successive Random Additions algorithm rather than the
Successive Random Displacement
; algorithm which Voss shows to be non-stationary.
;****** NOTE FUDGE NOTE FUDGE *******************
; The lines:
; IF i EQ 7 THEN Var = Var*0.9
; IF i EQ 8 THEN Var = Var*0.85
; IF i EQ 9 THEN Var = Var*0.5
; have been added in two places to correct the high frequency rise present
in the Voss spectrum
;****** NOTE FUDGE NOTE FUDGE *******************
nX = 2^N ; size of 2-D region modelled
X = FLTARR(nX+1,nX+1) ; this will hold the fractal surface
H = 1.0/3.0 ; dimension of surface is 3-H
Var = 1.0 ; variance of largest scale
STD = SQRT(Var)
L = 2^N ; L is the largest scale computer so far
LD2 = L/2 ; L/2 is the scale currently being computed
nL = 1
IDxL = [0]
IDxLD2 = [LD2]
IDxLPL = [L]
IDxLnX = [ 0 , L ]
FOR i = 0 , N-1 DO BEGIN
;*************************************************************
;** Stage 1: calculate the values at square centres **
;*************************************************************
; i.e we know A and are calculating B in A A
; B
; A A
; Do all the interior centre interpolation points
FOR iLy = 0 , nL-1 DO $
X[IDxLD2,LD2+iLy*L] = 0.25*( X[IDxL,iLy*L] + X[IDxLPL,iLy*L] +
X[IDxL,iLy*L+L] + X[IDxLPL,iLy*L+L] ) + STD*RANDOMN(SEED,nL)
; Update the previously calculated points
FOR iLy = 0 , nL DO $
X[IDxLnX,iLy*L] = X[IDxLnX,iLy*L] + STD*RANDOMN(SEED,nL+1)
Var = Var * SQRT(0.5)^(2.0*H) ; adjust variance
; IF i EQ 7 THEN Var = Var*0.9
; IF i EQ 8 THEN Var = Var*0.85
; IF i EQ 9 THEN Var = Var*0.5
STD = SQRT(Var)
;*****************************************************************
;** Stage 2: calculate the values around square centres **
;*****************************************************************
; i.e we know A and B and are calculating C in ACA
; CBC
; ACA
; The complexity comes from dealing with C's on the outer edge of the region
; Do all the boundary interpolation points
; Top and bottom
X[IDxLD2,0] = 0.5*( X[IDxL,0 ] + X[IDxLPL,0 ] ) + STD*RANDOMN(SEED,nL)
X[IDxLD2,nX] = 0.5*( X[IDxL,nX] + X[IDxLPL,nX] ) + STD*RANDOMN(SEED,nL)
; Left and right
X[0 ,IDxLD2] = 0.5*( X[0 ,IDxL] + X[0 ,IDxLPL] ) + STD*RANDOMN(SEED,nL)
X[nX,IDxLD2] = 0.5*( X[nX,IDxL] + X[nX,IDxLPL] ) + STD*RANDOMN(SEED,nL)
; Do all the C's inside the region
FOR iLy = 1 , nL-1 DO $
X[IDxLD2,iLy*L] = 0.25*( X[IDxL,iLy*L] + X[IDxLPL,iLy*L] +
X[IDxLD2,iLy*L-LD2] + X[IDxLD2,iLy*L+LD2] ) + STD*RANDOMN(SEED,nL)
IF i GT 0 THEN BEGIN
IDxL = IDxL[1:nL-1]
FOR iLy = 0 , nL-1 DO $
X[IDxL,LD2+iLy*L] = 0.25*( X[IDxL,iLy*L] + X[IDxL,iLy*L+L] +
X[IDxL-LD2,LD2+iLy*L] + X[IDxLD2,LD2+iLy*L] ) + STD*RANDOMN(SEED,nL)
ENDIF
;*****************************************************
;** Halve the scale for the next iteration **
;*****************************************************
L = L/2
LD2 = LD2/2
nL = nL*2
IDxL = INDGEN(nL)*L
IDxLD2 = IDxL + LD2
IDxLPL = IDxL + L
IDxLnX = [ IDxL , nX ]
; Update the previously calculated points
FOR iLy = 0 , nL DO $
X[IDxLnX,iLy*L] = X[IDxLnX,iLy*L] + STD*RANDOMN(SEED,nL+1)
Var = Var * SQRT(0.5)^(2.0*H) ; adjust variance
; IF i EQ 7 THEN Var = Var*0.9
; IF i EQ 8 THEN Var = Var*0.85
; IF i EQ 9 THEN Var = Var*0.5
STD = SQRT(Var)
ENDFOR
RETURN, X
END
------------------------