Geometric Interpretation of the Multi-solution Phenomenon in the P3P Problem

Abstract

It is well known that the P3P problem could have 1, 2, 3 and at most 4 positive solutions under different configurations among its three control points and the position of the optical center. Since in any real applications, the knowledge on the exact number of possible solutions is a prerequisite for selecting the right one among all the possible solutions, and the study on the phenomenon of multiple solutions in the P3P problem has been an active topic since its very inception. In this work, we provide some new geometric interpretations on the multi-solution phenomenon in the P3P problem, and our main results include: (1) the necessary and sufficient condition for the P3P problem to have a pair of side-sharing solutions is the two optical centers of the solutions both lie on one of the three vertical planes to the base plane of control points; (2) the necessary and sufficient condition for the P3P problem to have a pair of point-sharing solutions is the two optical centers of the solutions both lie on one of the three so-called skewed danger cylinders;(3) if the P3P problem has other solutions in addition to a pair of side-sharing (point-sharing) solutions, these remaining solutions must be a point-sharing (side-sharing ) pair. In a sense, the side-sharing pair and the point-sharing pair are companion pairs; (4) there indeed exist such P3P problems that have four completely distinct solutions, i.e., the solutions sharing neither a side nor a point, closing a long guessing issue in the literature. In sum, our results provide some new insights into the nature of the multi-solution phenomenon in the P3P problem, and in addition to their academic value, they could also be used as some theoretical guidance for practitioners in real applications to avoid occurrence of multiple solutions by properly arranging the control points.

Appendices

Appendix

Proof of four different P3P solutions by Macaulay 2

We use the coordinates of the three control points A, B, C and an optical center O as the original variables, and we set these parameters as rational number:

$$\begin{aligned} A&=[x_A, y_A, z_A]=[0, 0, 0], \\ B&=[x_B, y_B, z_B ]=[4, 5, 0], \\ C&=[x_C, y_C, z_C]=[10, 0, 0], \\ O&=[x_O, y_O, z_O ]=[11, -2, 4]. \end{aligned}$$

$|AB|, |BC|, |CA|, \cos \alpha , \cos \beta$ and $\cos \gamma$ can be expressed by the coordinates A, B, C and O as:

$$\begin{aligned} |AB|^2&=(x_A-x_B)^2+(y_A-y_B)^2+(z_A-z_B)^2\\ |BC|^2&=(x_B-x_C )^2+(y_B-y_C )^2+(z_B-z_C )^2\\ |CA|^2&=(x_C-x_A)^2+(y_C-y_A)^2+(z_C-z_A)^2\\ |OA|^2&=(x_O-x_A)^2+(y_O-y_A)^2+(z_O-z_A)^2\\ |OB|^2&=(x_O-x_B)^2+(y_O-y_B)^2+(z_O-z_B)^2\\ |OC|^2&=(x_O-x_C)^2+(y_O-y_C)^2+(z_O-z_C)^2\\ \cos \alpha&= (|OB|^2+|OC|^2-|BC|^2)/2|OB||OC|\\ \cos \beta&= (|OC|^2+|OA|^2-|CA|^2)/2|OC||OA|\\ \cos \gamma&= (|OA|^2+|OB|^2-|AB|^2)/2|OA||OB| \end{aligned}$$

Then, by substituting the above |AB|, |BC|, |CA|, $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ into Eq. (1), with $s_1, s_2, s_3$ as the variables. Then, an ideal is generated by the three equations in Eq. (1) over the field ${\mathbb {Q}}$(rational number). Then, we use the function “degree” in Macaulay to get the degree of this ideal. The output degree is 8, it means that there are eight different intersecting points of the three curves in Eq. (1). Since if $(s_1, s_2, s_3)$ satisfies Eq. (1), $-(s_1, s_2, s_3)$ also satisfies Eq. (1), and hence, it shows there exist four different P3P solutions.