四边形的对扭变换

zhouguang

1. lc[p1_, p2_] := Module[{x, y, p, c}, p = {x, y}; c = Flatten[CoefficientList[(p - p1).(p - p1) - (p - p2).(p - p2), p]]; c /= First[c]];
   
2. cp[p0_, p1_, p2_] := Module[{x, y, p, c}, p = {x, y}; c = p /. First[Solve[lc[p1, p2] - lc[p0, p] == 0, p]]; FullSimplify[c]];
   
3. pl = {-1, 0}; pr = {1, 0};
   
4. p1 = {x1, y1};
   
5. p2 = {x2, y2};
   
6. p3 = cp[p1, pl, p2]
   
7. p4 = cp[p2, pr, p3]
   
8. p5 = cp[p3, pl, p4]
   
9. p6 = cp[p4, pr, p5]
   
10. p7 = cp[p5, pl, p6]
    
11. p8 = cp[p6, pr, p7]

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函数lc用来求点p1、p2中垂线直线方程的系数，函数cp用来求p0关于p1p2中垂线的对称点。

mathe

数值计算好像是对于一般的四边形6次迭代复原

mathe

比如四边形
[1, 1.3000000000000000000000000000000000000, 1.7000000000000000000000000000000000000, 1.4587959601351103358923061605505501079, 1.5000000000000000000000000000000000000, 1.6000000000000000000000000000000000000, 1.0397200545554471787917385872838136848, 2.1434652526241392981335481792751920836] （其四个数据一次是边长，后四个数据依次是四个夹角的弧度）
然后6次变换结果依次为:
? onetran(%)
%78 = [1.7000000000000000000000000000000000000, 1.3000000000000000000000000000000000000, 1.4587959601351103358923061605505501079, 1, 1.6000000000000000000000000000000000000, 1.2972773348711535753827651780589964772, 2.1434652526241392981335481792751920836, 1.2424427196842936034089734092248172076]
? onetran(%)
%79 = [1.4587959601351103358923061605505501079, 1.3000000000000000000000000000000000000, 1, 1.7000000000000000000000000000000000000, 1.2972773348711535753827651780589964772, 2.2950357063922139203514693295395966337, 1.2424427196842936034089734092248172076, 1.4484295462319253777820788497355954499]
? onetran(%)
%80 = [1, 1.3000000000000000000000000000000000000, 1.7000000000000000000000000000000000000, 1.4587959601351103358923061605505501079, 2.2950357063922139203514693295395966337, 1.1250532024294051975121258426563723027, 1.4484295462319253777820788497355954499, 1.4146668521260419812796127446274413821]
? onetran(%)
%81 = [1.7000000000000000000000000000000000000, 1.3000000000000000000000000000000000000, 1.4587959601351103358923061605505501079, 1, 1.1250532024294051975121258426563723027, 1.8652496224799210815751794633538891228, 1.4146668521260419812796127446274413821, 1.8782156301442182165583687159213029608]
? onetran(%)
%82 = [1.4587959601351103358923061605505501079, 1.3000000000000000000000000000000000000, 1, 1.7000000000000000000000000000000000000, 1.8652496224799210815751794633538891228, 1.5000000000000000000000000000000000000, 1.8782156301442182165583687159213029608, 1.0397200545554471787917385872838136848]
? onetran(%)
%83 = [1, 1.3000000000000000000000000000000000000, 1.7000000000000000000000000000000000000, 1.4587959601351103358923061605505501079, 1.5000000000000000000000000000000000000, 1.6000000000000000000000000000000000000, 1.0397200545554471787917385872838136848, 2.1434652526241392981335481792751920836]

mathe

6#中对应的全等四边形染色后的结果，可以看出方向完全不同

zhouguang

每次扭转的对称轴都会交于一点，所以扭转操作相当于在置换这些三角形。

zhouguang

每次扭转的对称轴都会交于一点，所以扭转操作相当于在置换这些三角形。

yigo

hujunhua 发表于 2021-2-25 17:33
不太可能吧。

最近没有时间细心研究它，偷懒提到论坛想捡现成，要是这么简单，有点打脸啊。

由于镜像对称视为相同，对角线上的两点的任意其中一点变换是相同的，这样总可以保持一条边不动，经过3次变换，顶点就重合了，6楼的应该是变换中没有固定一条边不动。

zhouguang

图中显示了2次对扭变换。经过6次变换（3次交替）形状就会恢复。经过12次变换，位置也会恢复。

zhouguang

依次扭转ABCDA'B'之后，形状就会恢复，只是字母标号和位置都变化了。

mathe

楼上改变了变换的方式了，和胡氏变换不等价了