From: kib2 <kib2@fr...>  20061211 21:16:05

Thanks for your answers, the only problem I can see for the moment in the algorithm is that it=20 uses the Numpy package for matrix computation and for solving a linear=20 algebra system. ( with 2xn lines, where n is the initial number of=20 points ). Calculus are made with complex numbers. Maybe that could be a=20 problem if you wanted to put it inside PyX. It was just for testing my understanding of the Hobby's article. In fact=20 I don't have it, I asked on the list fr.comp.text.tex (=20 http://groups.google.fr/group/fr.comp.text.tex/browse_thread/thread/b0963= fc8e430a22b/?hl=3Dfr#=20 ) if someone had a clue and I got the chance that J.C.Charpentier=20 studied it a little before. He then explained to me the basics, because=20 he doesn't remind all the formulas with the tensions and other options. Now, the bad part :  I found the Python's complex module rather light : isn't there=20 anything to calculate the argument ( sorry, I don't know the English=20 equivalent ) of a complex ? ( I've written my own function for this, but=20 I may miss something I believe). It's the same for the exponential form=20 of a complex "e^(j*theta) =3D cos(theta) + j*sin(theta)" .  I don't know Numpy at all, I just started to use it yesterday. So=20 maybe the code looks ugly, the only reference I had is a table here :=20 http://www.scipy.org/NumPy_for_Matlab_Users  I saw a bug once : when points are "badly" placed, the rendering is=20 rather strange. But J.C.Charpentier told me it can be like this in=20 MetaPost too, in fact I don't know. I'll test this week, but if some of=20 you use MetaPost from time to time, maybe we can compare our results ? I'll try to do my best to post all this stuff tomorrow, see you : Kib=B2 Here's a little version for 3 points, don't care about the geoObject.=20 p0, p1, p2 are just Points instance with x and y attributes. Sorry, some comments are still in French but I don't have much time tonig= ht. # La fonction f de John Hobby: def hobby(t, p): sq2, sq5 =3D math.sqrt(2), math.sqrt(5) sint, sinp =3D math.sin(t), math.sin(p) cost, cosp =3D math.cos(t), math.cos(p) k=3D 1.0/16.0 num =3D 2 + sq2*( sint  k*sinp)*(sinp  k*sint)*(cost  cosp) den =3D 3*( 1 + 0.5*(sq51)*cost + 0.5*(3sq5)*cosp ) return float(num/den) # Complexe de la forme z =3D e^(i*theta) def e(theta): return complex( math.cos(theta) , math.sin(theta) ) =20 # Calcule l'argument d'un complexe z donne def arg(z): # Computes the argument (in radians) of a given complex number rez,imz =3D z.real, z.imag if rez =3D=3D 0.0: # on est dur l'axe des ordonnees if imz >0: return math.pi/2.0 if imz <0: return math.pi/2.0 if imz =3D=3D 0.0: return "Argument non defini pour un complexe nul \n" =20 truc =3D math.atan(imz/rez) if rez > 0.0: return truc if rez < 0: return truc + math.pi class HobbyBezier( pyx.path.path, geoObject ): """ Trace une courbe de Bezier passant par 3 Points donnes. =20 L'algorithme est de John Hobby dans le MetaFontBook. """ def __init__( self, p0, p1, p2): import numpy from numpy import linalg q =3D pyx.path.path() # z[] liste de complexes, on en compte n # pour z0..z1..z2 (n=3D2) z =3D [ complex(p0.x,p0.y) , complex(p1.x,p1.y) , complex(p2.x,p2= .y) ] # On decoupe l'intervalle de temps t en n morceaux l =3D [] # l[k] =3D distance entre z[k1] et z[k] (pour i dans [1= ,n]) =20 psi =3D [] # psi[k] =3D arg((z[k+1]z[k])/(z[k]z[k1])) (pour i=20 dans [1,n1], ici dans [1,2]) psi.append(0) psi.append( arg((z[2]z[1])/(z[1]z[0])) ) # k varie de 0 =E0 n1 si n est le nombre de points donnes l.append( 0 ) for k in range(2): k +=3D1 l.append( abs( z[k]z[k1]) ) print "longueurs =3D",l[1],l[2] theta =3D [] # n inconnues phi =3D [] # n inconnues =20 a =3D=20 numpy.mat([[0.,1.,1.,0.],[l[2],2.*l[1],2.*l[2],l[1]],[1.,0.,1.,0.],[0.= ,1.,0.,1.]]) print "a=3D\n",a,"\n" b =3D numpy.mat([ [psi[1]],[0.],[0.],[0.] ]) sol =3D linalg.solve(a,b) phi.append(0) # phi[0] does not count for i in range(4): if i < 2: theta.append( sol[i,0] ) else: phi.append( sol[i,0] ) =20 # Control Points u1 =3D z[0] + e(theta[0])*(z[1]z[0])*hobby( theta[0], phi[1] ) v1 =3D z[1]  e(phi[1])*(z[1]z[0])*hobby( phi[1], theta[0] ) u2 =3D z[1] + e(theta[1])*(z[2]z[1])*hobby( theta[1], phi[2] ) v2 =3D z[2]  e(phi[2])*(z[2]z[1])*hobby( phi[2], theta[1] ) =20 self.liste_controles =3D [ Point(u1.real, u1.imag), Point(v1.real= ,=20 v1.imag), Point(u2.real, u2.imag), Point(v2.real, v2.imag) ] =20 q.append( pyx.path.moveto( z[0].real, z[0].imag ) ) q.append( pyx.path.curveto( u1.real, u1.imag, v1.real, v1.imag,=20 z[1].real, z[1].imag ) ) q.append( pyx.path.curveto( u2.real, u2.imag, v2.real, v2.imag,=20 z[2].real, z[2].imag ) ) q.normpath( epsilon=3DNone ).path() pyx.path.path.__init__( self, *q.pathitems ) geoObject.__init__( self ) =20 def control_points( self ): return self.liste_controles =20 