Mathematica程序 计算两线段

1、Mathematica程序计算两线段之间的最短距离（*输入四个点坐标*）（*将坐标输成7/2形式可以得到准确解，输成3.5只能得到有一定精确度 的数值解.*）（*如果两线段分别为AB,CD,按照以下规则输入坐 标：xA,yA,xB,yB,xC,yC,xD,yD或者xA,yA,xB,yB,xC,yC,xD,yD或者两种混和输入，必须保证按 照顺序输入*）zuobiao = 0,0, 13,9, 6,6, 0, 8;zuobiao = PartitionFlattenzuobiao, 2;（*下面进行坐标旋转，目的是让两条直线均不与坐标轴平行，省去以后 很多分情况的麻烦，一劳永逸*）xuanzj

2、= 0, Pi/6, Pi/4;brr = ;err = ;juli = ;frr = ;chuizuf =.;Fori = 1, i = Lengthxuanzj, i+,Forj = 1, j =0 & (brr31 - brr51)*(brr51- brr41) = 0,Print-两线段相交，最短距离为0;xiangjiao = 1,xiangjiao = 0;(*下面先定义一个函数chuizu用来计算点到直线的垂足坐标*)chuizuzuobiao_:=Modulearr,arr = Flattenzuobiao;(-(-arr1 arr3A2 + 2 arr1 arr3 arr5-

3、arr1 arr5A2 - arr3 arr2 arr4 +arr5 arr2 arr4 - arr5 arr4A2 +arr3 arr2 arr6 - arr5 arr2 arr6 +arr3 arr4 arr6 + arr5 arr4 arr6-arr3 arr6A2)/(arr3A2 - 2 arr3 arr5 +arr5A2 + arr4A2 - 2 arr4 arr6 +arr6A2), -(-arr1 arr3 arr4 +arr1 arr5 arr4 + arr3 arr5 arr4-arr5A2 arr4 - arr2 arr4A2 +arr1 arr3 arr6 - arr3

4、A2 arr6-arr1 arr5 arr6 + arr3 arr5 arr6 +2 arr2 arr4 arr6-arr2 arr6A2)/(arr3A2 - 2 arr3 arr5 + arr5A2 + arr4A2 - 2 arr4 arr6 + arr6A2)；(*下面是两线段线不相交时，计算各个端点到所对线段的最短距离9crr = brr1, brr3, brr4, brr2, brr3,brr4, brr3, brr1, brr2, brr4, brr1, brr2;Ifxiangjiao = 0,Fori = 1, i =0 & (crri31 - chuizuf1)/(crri

5、31- crri21) = 0,frr = Appendfrr, crri1, chuizuf;juli =Appendjuli, SimplifyNormcrri1 - chuizuf(*A勺垂足 F 在CD内，取线段AF距离*);If(crri 21 - chuizuf1)/(crri21- crri31) = 0,frr = Appendfrr, crri1, crri 2 ;juli =Appendjuli,SimplifyNormcrri1 - crri2(*A的垂足F在C 的 外侧，取线段AC距离*);crri31) =0 & (crri31 - chuizuf1)/(crri31

6、- crri21) 0,frr = Appendfrr, crri1, crri3;juli =Appendjuli,SimplifyNormcrri1 - crri3(*A勺垂足F 在D 的 外侧，取线段AD距离*);Print准确解;PrintSortjuli, Less1;Print数值解;PrintSortjuli / N1;(*下面是作图*)Forj = 1, j = Lengthbrr, j+,err = Appenderr, brrj1*CosAlpha +brrj2*SinAlpha, -brrj1*SinAlpha +brrj2*CosAlpha;err = Partitionerr, 2;Iffrr != ,grr = TableNormfrri1 - frri 2 / N, i, 1, Lengthfrr;Fori = 1,