# Find rigid transformation with noisy data, simple approach

I have two set of points $$\left\{ a_j \right\}_{j=1...n},\left\{ b_j \right\}_{j=1...n}$$, you can assume $$a_j$$ are noisy. I want to find a rotation matrix $$R$$ and a translation vector $$T$$ such that

$$\epsilon(R,T) = \frac{1}{2n} \sum_{j=1}^{n} \lVert Ra_j + T - b_j \rVert_2^2$$

is minimized. I'm not sure what is the simplest approach, but here is what I was thinking, which I think make sense. I set $$R_0 = I$$ and $$T_0 = 0$$ (or maybe something more sensibile, exploiting also the content of this question per each iteration I alternate between minimizing w.r.t. $$T$$ (which is essentially a mean to be computed) and for finding $$R$$ at the current iteration I use the method linked in the wiki of the linked question (namely this).

Does this approach make sense? I don't like the "alternate" bit, but I can't see any other way (I can parametrize using quaternions, but I doubt it would be any simpler).

• Are there outliers ? – Yves Daoust Jan 22 '19 at 17:54
• You mean values with no correspondences? – user8469759 Jan 22 '19 at 17:57
• No, outliers are erratic points that do not belong to the true distribution. – Yves Daoust Jan 22 '19 at 17:58
• No, no outliers. – user8469759 Jan 22 '19 at 17:58

$$\epsilon(R) = \frac{1}{2n} \sum_{j=1}^{n} \lVert R\tilde a_j - \tilde b_j \rVert_2^2$$
This problem is not so easy, as the matrix $$R$$ is constrained to be orthogonal. If you have enough points, you can "cheat" by ignoring the constraint, solving the linear system, and ajusting orthogonality by Gram-Schmidt.
• How do I find $R$? Shall I use at that point the method I've pointed out? – user8469759 Jan 22 '19 at 18:00