/* Solving the associated legendre polynomials iteratively. Most of this code was taken from * the work described by Robin Green from Sony Computer Entertainment America: * * R. Green, Spherical Harmonic Lighting: The Gritty Details, Sony Computer Entertainment America, 2003 * http://www.research.scea.com/gdc2003/spherical-harmonic-lighting.pdf */ /// This function computes the factorial n! /// where n! = n * (n-1) * (n-2) * (n-3) * ...... * 2 * 1 unsigned long factorial(unsigned int n) { if (n == 0) //the standard definition of a factorial specifies that 0! = 1 return 1; return n * factorial(n - 1); //recursively call this function } /// Evaluate an Associated Legendre Polynomial P(l,m,x) at x /// This implements evaluation of the associated Legendre polynomials using /// a recursive method with the following rules: /// /// 1) (l - m)P(l, m) = x(2l-1)P(l-1, m) - (l + m - 1)P(l-2, m) /// 2) P(m, m) = (-1)^m (2m - 1)!! (1 - x^2)^{m/2} /// 3) P(m+1, m) = x (2m + 1)P(m, m) double P(int l,int m,double x) { double pmm = 1.0; //first iteration, P(0, 0) = 1 if(m>0) { //solve for positive values of m double somx2 = sqrt((1.0-x)*(1.0+x)); double fact = 1.0; for(int i=1; i<=m; i++) { pmm *= (-fact) * somx2; fact += 2.0; } } if(l==m) return pmm; //if the degree equals the order, the solution is found above double pmmp1 = x * (2.0*m+1.0) * pmm; if(l==m+1) return pmmp1; double pll = 0.0; for(int ll=m+2; ll<=l; ++ll) { pll = ( (2.0*ll-1.0)*x*pmmp1-(ll+m-1.0)*pmm ) / (ll-m); pmm = pmmp1; pmmp1 = pll; } return pll; } /// Calculate the scaling factor for a normalized Legendre polynomial /// l is the degree, ranging from [0..N] /// m is the order in the range [-l..l] double K(int l, int m) { static const double K_PI = 3.14159; // renormalisation constant for SH function double temp = ((2.0*l+1.0)*factorial(l-m)) / (4.0*K_PI*factorial(l+m)); return sqrt(temp); } /// return a point sample of a Spherical Harmonic basis function /// l is the degree, range [0..N] /// m is the order in the range [-l..l] /// theta in the range [0..Pi] /// phi in the range [0..2*Pi] double SH(int l, int m, double theta, double phi){ const double sqrt2 = sqrt(2.0); if(m==0) return K(l,0)*P(l,m,cos(theta)); else if(m>0) return sqrt2*K(l,m)*cos(m*phi)*P(l,m,cos(theta)); else return sqrt2*K(l,-m)*sin(-m*phi)*P(l,-m,cos(theta)); }