finding volume of an n-dimensional pyramid numerically

Question

In my experiment I need to compute hypervolume/area from a set of points,

let's start with a base case --

Triangle: In this case, I have 3 points in a 2D space and they make a triangle, $p_1 = \{x_1, y_1\}$, $p_2 = \{x_2, y_2\}$ and $p_3 = \{x_3, y_3\}$. Assume $p_3$ is the apex. The area of the triangle can be computed as $\frac{1}{2}|\mathbf{v_1} \times \mathbf{v_2}|$, where $\mathbf{v_1} = \{(x_2 - x_1),(y_2 - y_1)\}$ and $\mathbf{v_2} = \{(x_3 - x_1),(y_3 - y_1)\}$

Tetrahedron: In the case for 3D space, I will have 4 points, $p_1 = \{x_1, y_1, z_1\}$, $p_2 = \{x_2, y_2, z_2\}$, $p_3 = \{x_3, y_3, z_3\}$ and $p_4 = \{x_4, y_4, z_4\}$. $p_1$, $p_2$ and $p_3$ make up the base and $p_4$ is the apex. The volume can be computed as $\frac{1}{6} |\mathbf{v_1} \cdot (\mathbf{v_2} \times \mathbf{v_3})|$, where $\mathbf{v_1} = \{(x_4 - x_1),(y_4 - y_1),(z_4 - z_1)\}$, $\mathbf{v_2} = \{(x_4 - x_2),(y_4 - y_2),(z_4 - z_2)\}$ and $\mathbf{v_3} = \{(x_4 - x_3),(y_4 - y_3),(z_4 - z_3)\}$.

generalization --

N-dimension: In the case for n-dimensional space, I will have $1$ n-dimensional point as the apex and I will also have $n$ numbers of n-dimensional points that form the base (hyperplane). The solid that I need to form is an n-dimensional hyper-hedron (I am not sure if it's the right word) and I need to compute it's volume.

How do I generalize the (hyper-)volume computation for the n-dimensional hyper-hedron ?

Answer 1

I think you are looking for the volume of an $n$-dimensional simplex. If you have $n+1$ points in $n$ dimensions (where the points are given as vectors $v_0,v_1,\dots,v_{n}$), then the formula for the volume is given by

$$\left|\frac{1}{n!}\det\left[\begin{matrix}v_1-v_0\\v_2-v_0\\\vdots\\v_{n}-v_0\end{matrix}\right]\right|$$

Answer 2

Each dimension requires an adjusting factor because you're going for a "triangular" simplex instead of a "square" one. 1D Line segment is 1/1. Area of a 2D triangle is 1/2 the height times its base (line segment). Volume of a 3D tetrahedron is 1/3 the height times the base (triangle), which we just said was 1/2 of its height times its base. So we're up to 1/(3*2*1) = 1/(3!) already. Hypervolume of a 4D "hyper-tetrahedron" simplex is going to be 1/4 its height times its base (tetrahedron), which we just said was 1/6; so we're up to 1/(4!) so far. Etc.