最小二乘法拟合直线，对于大多数人都不是很陌生，直线方程 y=kx+b，令sum((y-kx-b)^2)，分别对k和b求导便可求出最佳的参数，但是如果遇到诸如“X=5”这样的直线方程，又该如何呢？这时候就不再适用了，参考这个方法，这样拟合出来的才是直线的一般方程。

Let the line sought be Ax + By + C = 0, with no restriction on A, B,
and C, except not both of A and B are zero. Then the distance from a
point (x[n],y[n]) to this line is given by

d[n] = |A*x[n]+B*y[n]+C|/sqrt(A^2+B^2).

Then the sum of the squares of the distances of the r points from the
line is

F(A,B,C) = SUM d[n]^2
         = SUM (A*x[n]+B*y[n]+C)^2/(A^2+B^2).

Here and henceforth, the sums run over the range 1 <= n <= r.

Now take the partial derivatives of F with respect to A, B, and C, and
set them equal to zero. This will give you a set of three simultaneous
nonlinear equations in A, B, and C, and you need to solve them. They
will be polynomial, of degree 3. When I did this, the equations turned
out to be

   SUM (A*x[n]+B*y[n]+C)*(A*C-B^2*x[n]+A*B*y[n]) = 0,
   SUM (A*x[n]+B*y[n]+C)*(B*C+A*B*x[n]-A^2*y[n]) = 0,
   SUM A*x[n]+B*y[n]+C = 0.

These can be rewritten as polynomials in A, B, and C with coefficients
involving

   r = SUM 1,
   s = SUM x[n],
   t = SUM y[n],
   u = SUM x[n]^2,
   v = SUM x[n]*y[n],
   w = SUM y[n]^2,

which are quantities you can compute from your data points.

   -v*A^2*B-s*A^2*C+(u-w)*A*B^2-2*t*A*B*C-r*A*C^2+v*B^3+s*B^2*C = 0,
   v*A^3+(w-u)*A^2*B+t*A^2*C-v*A*B^2-2*s*A*B*C-t*B^2*C-r*B*C^2 = 0,
   s*A+t*B+r*C = 0.

C can be eliminated using the last equation, which is the same as

       C = -(s*A+t*B)/r.

That will leave you with two cubic equations in two unknowns A and B.

They turn out to be equivalent to the single quadratic equation

   (r*v-s*t)*A^2 + (s^2-t^2-r*u+r*w)*A*B - (r*v-s*t)*B^2 = 0.
   A^2 + [(s^2-t^2-r*u+r*w)/(r*v-s*t)]*A*B - B^2 = 0.

Now unless rv - st = 0, you can solve this equation for -A/B, the
slope of the line. Then you can find C/B, too, and thus the equation
of the line. You will get two distinct solutions. One is the minimum
you seek, and the other is a saddle point of F(A,B,C). You can tell
which is the minimum by evaluating F(A,B,C) at each one and choosing
the solution that gives you the smaller value.

If rv = st, but the coefficient s^2-t^2-ru+rw is nonzero, then the
minimum will be achieved for either A = 0 or B = 0. Figure out C for
both of these possibilities, and substitute them into F(A,B,C). The
one that gives you the lesser value is your solution.

If rv = st and s^2-t^2-ru+rw = 0, then any line through the center
of mass (s/r,t/r) will give the same value of F as any other, and all
are tied for being the best.