论坛 › 算法交流 › 关于两个对应的非刚体变换的2D点集的匹配，使用Thin Plate Spline,求助

dianyancao 发表于 2012-5-21 22:01:58

(m13+m11*ax1+m12*ay1-bx1)^2+(m13+m11*ax2+m12*ay2-bx2)^2+(m13+m11*ax3+m12*ay3-bx3)^2+(m13+m11*ax4+m12*ay4-bx4)^2+(m23+m21*ax1+m22*ay1-by1)^2+(m23+m21*ax2+m22*ay2-by2)^2+(m23+m21*ax3+m22*ay3-by3)^2+(m23+m21*ax4+m22*ay4-by4)^2
http://zh.numberempire.com/cgi-bin/render2.cgi?%5Cnocache%20%5CLARGE%204%5C%2C%7B%5Cit%20m_%7B23%7D%7D%5E2%2B%5Cleft%28%5Cleft%282%5C%2C%7B%5Cit%20ay_4%7D%2B2%5C%2C%7B%5Cit%20ay_3%7D%2B2%5C%2C%20%7B%5Cit%20ay_2%7D%2B2%5C%2C%7B%5Cit%20ay_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B22%7D%7D%2B%5Cleft%282%5C%2C%7B%5Cit%20ax_4%7D%2B2%20%5C%2C%7B%5Cit%20ax_3%7D%2B2%5C%2C%7B%5Cit%20ax_2%7D%2B2%5C%2C%7B%5Cit%20ax_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B21%7D%7D-2%5C%2C%20%7B%5Cit%20by_4%7D-2%5C%2C%7B%5Cit%20by_3%7D-2%5C%2C%7B%5Cit%20by_2%7D-2%5C%2C%7B%5Cit%20by_1%7D%5Cright%29%5C%2C%20%7B%5Cit%20m_%7B23%7D%7D%2B%5Cleft%28%7B%5Cit%20ay_4%7D%5E2%2B%7B%5Cit%20ay_3%7D%5E2%2B%7B%5Cit%20ay_2%7D%5E2%2B%7B%5Cit%20ay_1%7D%20%5E2%5Cright%29%5C%2C%7B%5Cit%20m_%7B22%7D%7D%5E2%2B%5Cleft%28%5Cleft%282%5C%2C%7B%5Cit%20ax_4%7D%5C%2C%7B%5Cit%20ay_4%7D%2B2%5C%2C%20%7B%5Cit%20ax_3%7D%5C%2C%7B%5Cit%20ay_3%7D%2B2%5C%2C%7B%5Cit%20ax_2%7D%5C%2C%7B%5Cit%20ay_2%7D%2B2%5C%2C%7B%5Cit%20ax_1%7D%5C%2C%20%7B%5Cit%20ay_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B21%7D%7D-2%5C%2C%7B%5Cit%20ay_4%7D%5C%2C%7B%5Cit%20by_4%7D-2%5C%2C%20%7B%5Cit%20ay_3%7D%5C%2C%7B%5Cit%20by_3%7D-2%5C%2C%7B%5Cit%20ay_2%7D%5C%2C%7B%5Cit%20by_2%7D-2%5C%2C%7B%5Cit%20ay_1%7D%5C%2C%20%7B%5Cit%20by_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B22%7D%7D%2B%5Cleft%28%7B%5Cit%20ax_4%7D%5E2%2B%7B%5Cit%20ax_3%7D%5E2%2B%20%7B%5Cit%20ax_2%7D%5E2%2B%7B%5Cit%20ax_1%7D%5E2%5Cright%29%5C%2C%7B%5Cit%20m_%7B21%7D%7D%5E2%2B%5Cleft%28-2%5C%2C%20%7B%5Cit%20ax_4%7D%5C%2C%7B%5Cit%20by_4%7D-2%5C%2C%7B%5Cit%20ax_3%7D%5C%2C%7B%5Cit%20by_3%7D-2%5C%2C%7B%5Cit%20ax_2%7D%5C%2C%20%7B%5Cit%20by_2%7D-2%5C%2C%7B%5Cit%20ax_1%7D%5C%2C%7B%5Cit%20by_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B21%7D%7D%2B4%5C%2C%20%7B%5Cit%20m_%7B13%7D%7D%5E2%2B%5Cleft%28%5Cleft%282%5C%2C%7B%5Cit%20ay_4%7D%2B2%5C%2C%7B%5Cit%20ay_3%7D%2B2%5C%2C%7B%5Cit%20ay_2%7D%20%2B2%5C%2C%7B%5Cit%20ay_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B12%7D%7D%2B%5Cleft%282%5C%2C%7B%5Cit%20ax_4%7D%2B2%5C%2C%20%7B%5Cit%20ax_3%7D%2B2%5C%2C%7B%5Cit%20ax_2%7D%2B2%5C%2C%7B%5Cit%20ax_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B11%7D%7D-2%5C%2C%20%7B%5Cit%20bx_4%7D-2%5C%2C%7B%5Cit%20bx_3%7D-2%5C%2C%7B%5Cit%20bx_2%7D-2%5C%2C%7B%5Cit%20bx_1%7D%5Cright%29%5C%2C%20%7B%5Cit%20m_%7B13%7D%7D%2B%5Cleft%28%7B%5Cit%20ay_4%7D%5E2%2B%7B%5Cit%20ay_3%7D%5E2%2B%7B%5Cit%20ay_2%7D%5E2%2B%7B%5Cit%20ay_1%7D%20%5E2%5Cright%29%5C%2C%7B%5Cit%20m_%7B12%7D%7D%5E2%2B%5Cleft%28%5Cleft%282%5C%2C%7B%5Cit%20ax_4%7D%5C%2C%7B%5Cit%20ay_4%7D%2B2%5C%2C%20%7B%5Cit%20ax_3%7D%5C%2C%7B%5Cit%20ay_3%7D%2B2%5C%2C%7B%5Cit%20ax_2%7D%5C%2C%7B%5Cit%20ay_2%7D%2B2%5C%2C%7B%5Cit%20ax_1%7D%5C%2C%20%7B%5Cit%20ay_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B11%7D%7D-2%5C%2C%7B%5Cit%20ay_4%7D%5C%2C%7B%5Cit%20bx_4%7D-2%5C%2C%20%7B%5Cit%20ay_3%7D%5C%2C%7B%5Cit%20bx_3%7D-2%5C%2C%7B%5Cit%20ay_2%7D%5C%2C%7B%5Cit%20bx_2%7D-2%5C%2C%7B%5Cit%20ay_1%7D%5C%2C%20%7B%5Cit%20bx_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B12%7D%7D%2B%5Cleft%28%7B%5Cit%20ax_4%7D%5E2%2B%7B%5Cit%20ax_3%7D%5E2%2B%20%7B%5Cit%20ax_2%7D%5E2%2B%7B%5Cit%20ax_1%7D%5E2%5Cright%29%5C%2C%7B%5Cit%20m_%7B11%7D%7D%5E2%2B%5Cleft%28-2%5C%2C%20%7B%5Cit%20ax_4%7D%5C%2C%7B%5Cit%20bx_4%7D-2%5C%2C%7B%5Cit%20ax_3%7D%5C%2C%7B%5Cit%20bx_3%7D-2%5C%2C%7B%5Cit%20ax_2%7D%5C%2C%20%7B%5Cit%20bx_2%7D-2%5C%2C%7B%5Cit%20ax_1%7D%5C%2C%7B%5Cit%20bx_1%7D%5Cright%29%5C%2C%7B%5Cit%20m_%7B11%7D%7D%2B%7B%5Cit%20by_4%7D%20%5E2%2B%7B%5Cit%20by_3%7D%5E2%2B%7B%5Cit%20by_2%7D%5E2%2B%7B%5Cit%20by_1%7D%5E2%2B%7B%5Cit%20bx_4%7D%5E2%2B%7B%5Cit%20bx_3%7D%5E2%2B%20%7B%5Cit%20bx_2%7D%5E2%2B%7B%5Cit%20bx_1%7D%5E2

wayne 发表于 2012-5-22 17:08:49

好凌乱啊。
我都没了方向

dianyancao 发表于 2012-5-23 08:39:58

本帖最后由 dianyancao 于 2012-5-23 10:27 编辑

化简后的形式如下，c[]为已知数,求该式关于 {m11,m12,m13}的最小值不知道如何下手，:(凌乱中...
Minimize[c1*m11^2 + c2*m12^2 + c3*m13^2 + c4*m11*m12 + c5*m11*m13 +
c6*m12*m13 + c7*m11 + c8*m12 + c9*m13 + c10, {m11, m12,m13}]

dianyancao 发表于 2012-5-24 09:12:30

原来很简单，这个问题好像叫最小2乘解，直接配方就可以了
结果：http://img.my.csdn.net/uploads/201205/24/1337821629_4739.jpg

dianyancao 发表于 2012-5-24 09:16:48

附上Mathematica的代码f =
c1*m11^2 + c2*m12^2 + c3*m13^2 + c4*m11*m12 + c5*m11*m13 +
c6*m12*m13 + c7*m11 + c8*m12 + c9*m13 + c10

rt = {t1, t2, t3}
r = {x11, x12, x13}

{x11, x12, x13}







mm1 = CoefficientList, m11]

r[] = (mm1[]/(2*mm1[]))

rt[] = -(mm1[]/(2*mm1[]))^2*mm1[] + mm1[]

mm2 = CoefficientList], m12]

r[] = (mm2[]/(2*mm2[]))

rt[] = -(mm2[]/(2*mm2[]))^2*mm2[] + mm2[]

mm3 = CoefficientList], m13]

r[] = (mm3[]/(2*mm3[]))
rt[] = -(mm3[]/(2*mm3[]))^2*mm3[] + mm3[]

Solve[{m11 + r[] == 0, m12 + r[] == 0, m13 + r[] == 0}, {m11,
   m12, m13}]

dianyancao 发表于 2012-5-25 07:41:44

遇到一个新问题，如果是采用矩阵元 1-范数即取绝对值，得到的结果形如：
http://zh.numberempire.com/equation.render?\huge \sum_{i=1}^{n}{\mid x_i*m_1 + y_i*m_2 + m_3 - z_i\mid }
如何计算该式的最小值呢，用分类讨论貌似很低效

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