Triangulate 3D Point from Two Views

hardPROMultiple CamerasComputer Vision 1

Given two camera projection matrices $P_1$ (3x4) and $P_2$ (3x4), and corresponding 2D points $\mathbf{x}_1 = [u_1, v_1]$ and $\mathbf{x}_2 = [u_2, v_2]$ in each image, triangulate the 3D point $\mathbf{X} = [X, Y, Z]$ using the Direct Linear Transform (DLT) method.

Algorithm (DLT Triangulation):

For each 2D point $\mathbf{x} = [u, v]$ and its projection matrix $P$ , we have: $\mathbf{x} \times (P \cdot \mathbf{X}) = 0$

This gives us a system of linear equations. From each view, we extract two independent equations:

$u_i \cdot P_i^{3T} - P_i^{1T} = 0$ $v_i \cdot P_i^{3T} - P_i^{2T} = 0$

where $P_i^{jT}$ is the $j$ -th row of $P_i$ .

Build the $4 \times 4$ matrix $A$ :

$A = \begin{pmatrix} u_1 P_1^{3T} - P_1^{1T} \\ v_1 P_1^{3T} - P_1^{2T} \\ u_2 P_2^{3T} - P_2^{1T} \\ v_2 P_2^{3T} - P_2^{2T} \end{pmatrix}$

Solve via SVD of $A$ : the solution is the last column of $V$ (last row of $V^T$ ), converted from homogeneous coordinates.

Round each coordinate to 4 decimal places.

Example

Input

P1 = [[1, 0, 0, 0],
      [0, 1, 0, 0],
      [0, 0, 1, 0]]
P2 = [[1, 0, 0, -1],
      [0, 1, 0, 0],
      [0, 0, 1, 0]]
x1 = [0.5, 0.5]
x2 = [0.0, 0.5]

Output

[1.0, 1.0, 2.0]

Reasoning

Inputs: $P_1 = [I \mid 0]$ ; $P_2$ is the same camera shifted one unit along $x$ . The point is seen at $(0.5, 0.5)$ and $(0.0, 0.5)$ .
Rows of $A$ : with $P^{3T} = [0, 0, 1, 0]$ in both cameras, the four rows are $[-1, 0, 0.5, 0]$ , $[0, -1, 0.5, 0]$ , $[-1, 0, 0, 1]$ and $[0, -1, 0.5, 0]$ .
Solving $A\mathbf{X} = 0$ with $\mathbf{X} = [X, Y, Z, W]$ : row 1 gives $Z = 2X$ , row 2 gives $Y = Z/2$ , row 3 gives $W = X$ . Every solution is a multiple of $[1, 1, 2, 1]$ ; the unit-length one found is $\approx -0.378 \cdot [1, 1, 2, 1]$ .
Homogeneous to Euclidean: dividing the first three entries by $W$ cancels both the scale and the sign, giving [1.0, 1.0, 2.0].
Notice that rows 2 and 4 are identical (the cameras differ only along $x$ , so $v$ agrees), so $A$ has rank 3 and the null space is exact. With noisy points, $A$ has full rank and you get the best least-squares fit instead.

Constraints

P1, P2: 3x4 projection matrices as lists of lists
x1, x2: 2D points as [u, v]
Use numpy for SVD
Return: [X, Y, Z] rounded to 4 decimal places
The 3D point is in front of both cameras

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Background Knowledge

The problem of triangulating a 3D point from two views is a fundamental concept in computer vision, particularly in the field of multiple view geometry. It involves estimating the 3D coordinates of a point given its projections in two or more images taken from different viewpoints. This is achieved by exploiting the geometric relationships between the cameras, the 3D point, and its image projections. The Direct Linear Transform (DLT) method is a popular approach for solving this problem, which formulates the triangulation process as a system of linear equations.

The DLT method relies on the concept of homogeneous coordinates, where a 3D point $\mathbf{X} = [X, Y, Z]$ is represented as $\mathbf{X} = [X, Y, Z, 1]^T$ to facilitate linear transformations. The projection of $\mathbf{X}$ onto an image plane is described by a $3 \times 4$ projection matrix $P$ , which maps the 3D point to its 2D image coordinates $\mathbf{x} = [u, v]^T$ . The relationship between $\mathbf{x}$ , $P$ , and $\mathbf{X}$ is given by the equation $\mathbf{x} \times (P \cdot$ \mathbf{X} $) = 0$ , which provides the basis for deriving the system of linear equations used in the DLT method.

In the context of this problem, we are given two camera projection matrices $P_1$ and $P_2$ , along with the corresponding 2D points $\mathbf{x}_1$ and $\mathbf{x}_2$ . The goal is to triangulate the 3D point $\mathbf{X}$ using the DLT method, which involves building a $4 \times 4$ matrix $A$ from the given information and solving for $\mathbf{X}$ via Singular Value Decomposition (SVD). Understanding the geometric and algebraic principles underlying the DLT method is essential for tackling this problem.

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