github: https://github.com/devernay/cminpack
主页: http://devernay.github.io/cminpack/
使用手册: http://devernay.github.io/cminpack/man.htmlhtml
lmdif
,lmdif1_
- 最小化非线性函数平方和git
include <minpack.h> void lmdif1_(void (*fcn)(int *m, int *n, double *x, double *fvec, int *iflag), int *m, int *n, double *x, double *fvec, double *tol, int *info, int *iwa, double *wa, int *lwa); void lmdif_(void (*fcn)(int *m, int *n, double *x, double *fvec, int *iflag), int *m, int *n, double *x, double *fvec, double *ftol, double *xtol, double *gtol, int *maxfev, double *epsfcn, double *diag, int *mode, double *factor, int *nprint, int *info, int *nfev, double *fjac, int *ldfjac, int *ipvt, double *qtf, double *wa1, double *wa2, double *wa3, double *wa4 );
lmdif_
的目的是最小化m个n元非线性方程的平方和,使用的方法是LM算法的改进。用户须要提供计算方程的子程序。Jacobian矩阵会经过一个前向差分(forward-difference)近似计算获得。
lmdif1_
是相同的目的,可是调用方法更简单一些。github
这些函数是经过FORTRAN写的,若是从C调用,须要记住如下几点:算法
void fcn(int m, int n, double *x, double *fvec, int *iflag) { /* 计算函数在x点的值,经过fvec返回。*/ }iflag的值不能被fcn所修改,除非用户想要终止
lmdif
/lmdif1_
。在这个例子中iflag设置为负整数。m:函数个数;
n:变量个数(n<=m)
x:长度为n的数组,设置为初始的估计解向量。输出的时候x内容为最终估计的解向量。
fvec:输出长度为m的数组,内容为最终输出x计算获得的函数解。数组
tol:做为输入,非负数。用于函数终止的条件判断:app
info:做为输出。若是用户终止了函数的执行,info将被设置为iflag的值(负数)(详细见fcn的描述),不然,info的值以下几种状况:less
iwa:长度n的工做数组;
wa:长度lwa的工做数组;
lwa:做为输入,整数,不能小于mn+5n+m;
[NOTE] 这三个输入我也不知道做用,从样例来看不须要初始化。ide
暂时不用这部分,跳过。函数
/* driver for lmdif1 example. */ #include <stdio.h> #include <math.h> #include <assert.h> #include <minpack.h> #define real __minpack_real__ // 用户自定义的函数f() // real -> __cminpack_real__ -> 浮点数(double) void fcn(const int *m, const int *n, const real *x, real *fvec, int *iflag); int main() { int j, m, n, info, lwa, iwa[3], one=1; real tol, fnorm, x[3], fvec[15], wa[75]; // 函数个数15; 变量数3 m = 15; n = 3; /* the following starting values provide a rough fit. */ // 初始位置 // 1.e0 = 1.0e0 = 1.0 x[0] = 1.e0; x[1] = 1.e0; x[2] = 1.e0; // 为何要设置75? lwa = 75; /* set tol to the square root of the machine precision. unless high precision solutions are required, this is the recommended setting. */ // (建议打印一下看值是多少) tol = sqrt(__minpack_func__(dpmpar)(&one)); // 须要注意指针 __minpack_func__(lmdif1)(&fcn, &m, &n, x, fvec, &tol, &info, iwa, wa, &lwa); // 最终的2范数(即平方和开根号) fnorm = __minpack_func__(enorm)(&m, fvec); printf(" final l2 norm of the residuals%15.7g\n\n", (double)fnorm); printf(" exit parameter %10i\n\n", info); printf(" final approximate solution\n"); for (j=1; j<=n; j++) { printf("%s%15.7g", j%3==1?"\n ":"", (double)x[j-1]); } printf("\n"); return 0; } // The problem is to determine the values of x(1), x(2), and x(3) // which provide the best fit (in the least squares sense) of // x(1) + u(i)/(v(i)*x(2) + w(i)*x(3)), i = 1, 15 // to the data // y = (0.14,0.18,0.22,0.25,0.29,0.32,0.35,0.39, // 0.37,0.58,0.73,0.96,1.34,2.10,4.39), // where u(i) = i, v(i) = 16 - i, and w(i) = min(u(i),v(i)). The // i-th component of FVEC is thus defined by // y(i) - (x(1) + u(i)/(v(i)*x(2) + w(i)*x(3))). void fcn(const int *m, const int *n, const real *x, real *fvec, int *iflag) { /* function fcn for lmdif1 example */ int i; real tmp1,tmp2,tmp3; // 实际的y值 real y[15]={1.4e-1,1.8e-1,2.2e-1,2.5e-1,2.9e-1,3.2e-1,3.5e-1,3.9e-1, 3.7e-1,5.8e-1,7.3e-1,9.6e-1,1.34e0,2.1e0,4.39e0}; assert(*m == 15 && *n == 3); if (*iflag == 0) { /* insert print statements here when nprint is positive. */ /* if the nprint parameter to lmder is positive, the function is called every nprint iterations with iflag=0, so that the function may perform special operations, such as printing residuals. */ // 这段没有很看懂,在??状况下打印信息 return; } /* compute residuals */ for (i=0; i<15; i++) { tmp1 = i+1; tmp2 = 15 - i; tmp3 = tmp1; if (i >= 8) { tmp3 = tmp2; } fvec[i] = y[i] - (x[0] + tmp1/(x[1]*tmp2 + x[2]*tmp3)); } }
[NOTE] 其余内容有待更新