norm-one direction $\mathbf{d}$ and a point ${{\mathbf{y}}_{l}}$ on it as $$\label{equ:2}
{{\mathbf{r}}_{l}}=\left( {{\mathbf{I}}_{3}}-\mathbf{d}{{\mathbf{d}}^{T}} \right)\left( \mathbf{R}{{\mathbf{x}}_{l}}+\mathbf{t}-{{\mathbf{y}}_{l}} \right)$$ where ${{\mathbf{I}}_{3}}$ is the identify matrix. Lastly, a point-to-point correspondence yields a $3$-dimensional residual vector $$\label{equ:3}
{{\mathbf{r}}_{p}}=\mathbf{R}{{\mathbf{x}}_{p}}+\mathbf{t}-{{\mathbf{y}}_{p}}$$

$\mathbf{R}$ and $\mathbf{t}$ have 6 DoF. Our problem is to calculate $\mathbf{R}$ and $\mathbf{t}$ from any configuration of ${{n}_{\pi }}$ point-to-plane, ${{n}_{l}}$ point-to-line, and ${{n}_{p}}$ point-to-point correspondences that has at least 6 effective constraints. Let use define $N$ as the number of effective correspondences. When $N > 6$, this is a least squares problem described in section \[sec:ls\]. When $N=6$, this forms a minimal problem solved in section \[sec:min\].

Least Squares Solution {#sec:ls}
======================

Using the notation in (\[equ:1\]), (\[equ:2\]) and (\[equ:3\]), we define our cost function as follows $$\label{equ:4}
\begin{array}{l}
{C_{\pi lp}}\left({\bf{R}},{\bf{t}}\right) = \sum\limits_{i = 1}^{{n_\pi }} {r_{{\pi _i}}^2}  + \sum\limits_{i = 1}^{{n_l}} {{\bf{r}}_{{l_i}}^T{{\bf{r}}_{{l_i}}}}  + \sum\limits_{i = 1}^{{n_p}} {{\bf{r}}_{{p_i}}^T{{\bf{r}}_{{p_i}}}} ,\\
s.t.\;{\bf{R}}{{\bf{R}}^T} = {{\bf{I}}_3},\det \left( {\bf{R}} \right) = 1,
\end{array}$$ where $\det \left( \mathbf{R} \right)$ represents the determinant of $\mathbf{R}$.

Ambiguous Configurations
------------------------

In previous works, the global minimizer is treated as the optimal solution. It is well known that there is an unique solution for at least three point-to-point correspondences, except for the degenerate configuration, all 