" Obviously FindGeometricTransform yields a unit matrix for all n>=8."

This means that it the algorithm is producing nonsense even for small exact image pairs.

The procedure seems to be buggy in other aspects, too.

https://mathematica.stackexchange.com/questions/232750/findgeometrictransform-not-working-properly-for-a-simple-problem

Replace Det by MatrixForm and you observe that FindGeometricTransform is obviously giving a unit matrix for n>=8.

By the way only the homogenous matrix with both right and bottom margins removed has to have det =-1 for a reflection at a hyperplane.

Now, I try to further check your example code as follows:

In[101]:= n = 8;
PointSet = Partition[RandomReal[{-25, 25}, n 7], n];
ReflectedSet = 
  ReflectionTransform[r1 = RandomReal[{-5, 5}, n], 
    r2 = RandomReal[{-5, 5}, n]][
   Permute[PointSet, RandomPermutation[Length[PointSet]]]];
FindGeometricTransform[PointSet, ReflectedSet] // Last // 
  TransformationMatrix // Det

Out[104]= 1.

As we all know, a mirror reflection transformation must correspond to the matrix whose determinant is -1, which is inconsistent with the above result. Therefore, I am still puzzled by your approach. Any more hints will be appreciated.

Regards, Zhao

Thanks a lot for your nice example. I tried as follows but failed:

In[342]:= n = 8;
PointSet = Partition[RandomReal[{-25, 25}, 15], n];
center1 = Total[PointSet]/Length[PointSet];
ReflectedSet = 
  ReflectionTransform[r1 = RandomReal[{-5, 5}, n], 
    r2 = RandomReal[{-5, 5}, n]][
   Permute[PointSet, RandomPermutation[Length[PointSet]]]];
center2 = Total[ReflectedSet]/Length[ReflectedSet];
 mid = (center1 + center2)/2;
dir = center2 - center1;

ReflectionTransform[dir, mid] // TransformationMatrix // InputForm
ReflectionTransform[r1, r2] // TransformationMatrix // InputForm
% == %%
Out[351]= False
{{0.5099503864003545, 0.18538145813340623, -0.02469833515332397, -0.01995727838866815, -0.04645357867574137, -0.3024036518959173, -0.6739143575470887, 
0.39599185939935283, -2.1655063144913793}, {0.18538145813340623, 0.9298718250847475, 0.009343162931113624, 0.007549662861459084, 0.017572980186980138, 
0.11439664143706756, 0.25493587341379215, -0.14980023709269571, 0.8191919900287443}, {-0.02469833515332397, 0.009343162931113624, 0.9987552122429704, 
-0.0010058400960075197, -0.0023412446890369175, -0.015241042006445287, -0.033965056201295286, 0.019957856082350425, -0.1091407874788013}, {-0.01995727838866815, 
0.007549662861459084, -0.0010058400960075197, 0.9991872395169192, -0.0018918227380525364, -0.012315393582918402, -0.027445173040531144, 0.016126774837405846, 
-0.08819027945613422}, {-0.04645357867574137, 0.017572980186980138, -0.0023412446890369175, -0.0018918227380525364, 0.99559649694256, -0.028665937989403326, 
-0.06388278402888639, 0.037537503316121906, -0.20527619073937942}, {-0.3024036518959173, 0.11439664143706756, -0.015241042006445287, -0.012315393582918402, 
-0.028665937989403326, 0.8133903871319097, -0.4158643474695681, 0.2443617566062408, -1.3363075891349618}, {-0.6739143575470887, 0.25493587341379215, 
-0.033965056201295286, -0.027445173040531144, -0.06388278402888639, -0.4158643474695681, 0.0732355485966375, 0.5445664930959656, -2.9779960154950267}, 
{0.39599185939935283, -0.14980023709269571, 0.019957856082350425, 0.016126774837405846, 0.037537503316121906, 0.2443617566062408, 0.5445664930959656, 
0.6800129040839014, 1.7498694993708317}, {0., 0., 0., 0., 0., 0., 0., 0., 1.}}

Hi Hongyi,

the programming idea could be to follow path:

A nessecary conditon for two sets of points to be images of each other under the euclidean group is the equality of the sets of all pair distances. If all distances are different, the pairs can be ordered in both sets and the pairs wirh the same index are possible images of each other. The question is, if all images are produced by the same transformation.

For each mirror pair of four points there are two possible joining lines and two corresponding mirror hyperplanes though their midpoints. The set of these pairs of hyperplanes has to contain the same hyperplane for each mirror pair of pairs.

Regards Roland