编辑
这经历了很多修改和想法。为简单起见，我编辑了大部分内容，然后给你最终结果。
这非常适合您的功能，除了中间的一个区域（温度从 256 到 316）有一点偏差。
你的函数的问题是它有近似无限的解，所以要很好地解决它，你需要更多的约束，但是什么？ Ol Sen 的参考资料https://www.waveformlighting.com/tech/calculate-color-temperature-cct-from-cie-1931-xy-coordinates 详细讨论了它，然后提到您希望Duv 为零。它还提供了一种计算 Duv 的方法，因此我将其添加到我的优化器中，瞧！
漂亮而流畅。优化器现在求解 x 和 y，既满足您的功能又最小化 Duv。
为了让它很好地工作，我不得不对 Duv 进行相当大的扩展。该页面提到 Duv 应该非常小，所以我认为这是一件好事。此外，随着温度的升高，缩放应该有助于优化器。
下面的打印从 153 到 500。
#import <Foundation/Foundation.h>
// Function taken from your code
// Simplified a bit
int ctFuncI ( float x, float y )
{
// float R1 = [hue[0] floatValue];
// float S1 = [hue[1] floatValue];
float result = (x-0.332)/(y-0.1858);
float cubic = - 449 * result * result * result + 3525 * result * result - 6823.3 * result + 5520.33;
float micro2 = 1 / cubic * 1000000;
int ct = ( int )( micro2 + 0.5 );
if ( ct < 153 )
{
ct = 153;
}
return ct;
}
// Need this
// Float version of your code
float ctFuncF ( float x, float y )
{
// float R1 = [hue[0] floatValue];
// float S1 = [hue[1] floatValue];
float result = (x-0.332)/(y-0.1858);
float cubic = - 449 * result * result * result + 3525 * result * result - 6823.3 * result + 5520.33;
return 1000000 / cubic;
}
// We need an additional constraint
// https://www.waveformlighting.com/tech/calculate-duv-from-cie-1931-xy-coordinates
// Given x, y calculate Duv
// We want this to be 0
float duv ( float x, float y )
{
float f = 1 / ( - 2 * x + 12 * y + 3 );
float u = 4 * x * f;
float v = 6 * y * f;
// I'm typing float but my heart yells double
float k6 = -0.00616793;
float k5 = 0.0893944;
float k4 = -0.5179722;
float k3 = 1.5317403;
float k2 = -2.4243787;
float k1 = 1.925865;
float k0 = -0.471106;
float du = u - 0.292;
float dv = v - 0.24;
float Lfp = sqrt ( du * du + dv * dv );
float a = acos( du / Lfp );
float Lbb = k6 * pow ( a, 6 ) + k5 * pow( a, 5 ) + k4 * pow( a, 4 ) + k3 * pow( a, 3 ) + k2 * pow(a,2) + k1 * a + k0;
return Lfp - Lbb;
}
// Solver!
// Returns iterations
int ctSolve ( int ct, float * x, float * y )
{
int iter = 0;
float dx = 0.001;
float dy = 0.001;
// Error
// Note we scale duv a bit
// Seems the higher the temp, the higher scale we require
// Also note the jump at 255 ...
float s = 1000 * ( ct > 255 ? 10 : 1 );
float d = fabs( ctFuncF ( * x, * y ) - ct ) + s * fabs( duv ( * x, * y ) );
// Approx
while ( d > 0.5 && iter < 250 )
{
iter ++;
dx *= fabs( ctFuncF ( * x + dx, * y ) - ct ) + s * fabs( duv ( * x + dx, * y ) ) < d ? 1.2 : - 0.5;
dy *= fabs( ctFuncF ( * x, * y + dy ) - ct ) + s * fabs( duv ( * x, * y + dy ) ) < d ? 1.2 : - 0.5;
* x += dx;
* y += dy;
d = fabs( ctFuncF ( * x, * y ) - ct ) + s * fabs( duv ( * x, * y ) );
}
return iter;
}
// Tester
int main(int argc, const char * argv[]) {
@autoreleasepool
{
// insert code here...
NSLog(@"Hello, World!");
float x, y;
int sume = 0;
int sumi = 0;
for ( int ct = 153; ct <= 500; ct ++ )
{
// Initial guess
x = 0.4;
y = 0.4;
// Approx
int iter = ctSolve ( ct, & x, & y );
// CT and error
int ctEst = ctFuncI ( x, y );
int e = ct - ctEst;
// Diagnostics
sume += abs ( e );
sumi += iter;
// Print out results
NSLog ( @"want ct = %d x = %f y = %f got ct %d in %d iter error %d", ct, x, y, ctEst, iter, e );
}
NSLog ( @"Sum of abs errors %d iterations %d", sume, sumi );
}
return 0;
}
要使用它，请执行以下操作。
// To call it, init x and y to some guess
float x = 0.4;
float y = 0.4;
// Then call solver with your temp
int ct = 217;
ctSolve( ct, & x, & y ); // Note you pass references to x and y
// Done, answer now in x and y