Hongyi Zhao on Tue, 14 Mar 2023 10:44:16 +0100 |
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Re: Determine the mirror reflection relationship between the coordinates of two sets of pairs of points in n-dimension space. |
On Mon, Mar 13, 2023 at 5:19 PM Karim Belabas <Karim.Belabas@math.u-bordeaux.fr> wrote: > > * Hongyi Zhao [2023-03-13 04:08]: > > But in all the above steps, I can only see that there is exactly one v > > is found. > > (Up to normalization, yes.) > > > So, the complete solution set of this problem is still not obtained so > > far. > > It is: you get a (short) list V of potential v's; for each such v you get > a (short) list C of potential c's; and we know that the full solution set > is inside { s_{v,c}, v in V, c in C }. Where solutions attached to (a.v, a.c) > for a non-zero scalar 'a' have been grouped together by our normalization. > > In this particular case, a single (v,c) is found, and works. (And the > affine hyperplane it corresponds to is given by the equation <x,v> = c.) Now, let's discuss a further question as follows: Determine the affine transformation relationship between the coordinates of two sets of points in n-dimensional space. For this purpose, I've provided the following example data in 4-dimensional space: ? A=[ -x, x, x+1, x, x, x; -y, y+1/6, y, y+1, y, y; -z-t, -t, z, z, z+1, z; t, z+t, t, t, t, t+1 ]; ? B=[ -x, x+2/3, x+1, x, x, x; -y, y+1/6, y, y+1, y, y; -z-t, -t, z, z, z+1, z; t, z+t, t, t, t, t+1 ]; They represent two sets of points in 4-dimensional space, with 6 points in each set. Furthermore, I've known that these two set of points can be connected by an affine transformation matrix, such as following one: ? afftran=[ -1, -2, 0, 0, 0; 1, 1, 0, 0, -1/3; 0, 0, 0, 1, 0; 0, 0, 1, 0, 0; 0, 0, 0, 0, 1 ]; Now, the question is: how to determine such matrices from scratch only based on the above two sets of points? > Cheers, > > K.B. Regards, Zhao > -- > Pr Karim Belabas, U. Bordeaux, Vice-président en charge du Numérique > Institut de Mathématiques de Bordeaux UMR 5251 - (+33) 05 40 00 29 77 > http://www.math.u-bordeaux.fr/~kbelabas/ > `