uniformly sample a point in a triangle $(1,0,0), (0,1,0), (0,0,1)$

Question

To choose a point with uniform distribution in a triangle $A:(1,0,0), B:(0,1,0), C:(0,0,1)$, my thought is to project the triangle onto X-Y plane first, and the projected triangle is $A:(1,0,0), B:(0,1,0), C':(0,0,0)$. Then sample a point $(x,y,0)$ in the projected triangle. Finally transform the point back to the triangle in the 3D space and the coordinate is $(x,y,1-x-y)$.

However, I feel like there are some problems with this approach. The areas of these two triangles are different. Yet the points before and after the tranfor have the same X- and Y-coordinates. Is this correct?

Answer 1

Your method does work, because under the transformation $(x,y,0) \mapsto (x,y,1-x-y)$, areas remain proportional. So in particular, the area of the triangle $\mathcal T(t) = \{(0,0), (t,0), (0,t)\}$ for $0 \le t \le 1$ is $t^2/2$, and under transformation, the triangle is mapped to $\mathcal T'(t) = \{(0,0,1), (t,0,1-t), (0,t,1-t)\}$ which has area $t^2 \frac{\sqrt{3}}{2}$. So, provided that you select $(x,y)$ uniformly in the triangle $\mathcal T(1)$, then $(x,y,1-x-y)$ will be uniformly chosen in $\mathcal T'(1)$.

So the only remaining issue is how to choose such $(x,y)$ uniformly. This is quite easy, since the conditional density of $Y$ given $X = x \in [0,1]$ is $$f_{Y \mid X}(y \mid x) = \frac{1}{1-x} \mathbb 1 (0 \le y \le 1-x).$$ So in order for $(X,Y)$ to be uniform on $\mathcal T(1)$, we must choose $X \in [0,1]$ inversely proportional to the length of the support of $Y \mid X$; i.e., $$f_X(x) \propto 1-x,$$ hence $$f_X(x) = 2(1-x), \quad 0 \le x \le 1.$$ This is a $\operatorname{Beta}(1,2)$ distribution, so we can write $$X \sim \operatorname{Beta}(1,2) \\ Y \mid X \sim \operatorname{Uniform}(0, 1-X) \\ Z = 1 - X - Y.$$

The choice of $X$ is the most difficult part, but we can use the inverse transform sampling method because this particular density has an easily invertible CDF. We choose $U \sim \operatorname{Uniform}(0,1)$, then compute $X = 1 - \sqrt{1 - U}$, which will be $\operatorname{Beta}(1,2)$ distributed. Since $U$ is symmetric on $[0,1]$, we can eliminate one operation and just let $X = 1 - \sqrt{U}$.

An alternative approach is to compute $(X,Y)$ uniform on the square $[0,1]^2$, and then employ a conditional: if $X+Y > 1$, then let the selected point be $(1-X, 1-Y, X+Y-1)$, in effect rotating the point $(X,Y)$ about $(1/2, 1/2)$. So this algorithm looks like this: $$X' \sim \operatorname{Uniform}(0,1) \\ Y' \sim \operatorname{Uniform}(0,1) \\ (X,Y,Z) = \begin{cases}(X',Y',1-X'-Y'), & X' + Y' \le 1 \\ (1-X', 1-Y', X'+Y'-1), & X' + Y' > 1. \end{cases}$$ I'm not actually sure which approach is faster, but the square root in the first method may be more computationally expensive.

Answer 2

A more general answer, valid for any triangle.

A direct way to get a uniform distribution of points in a triangle $ABC$ is by using this barycentrical expression :

$$M=(1-\sqrt{r_{1}})A+(\sqrt{r_{1}}r_{2})B+\sqrt{r_{1}}(1-r_{2})C\tag{1}$$

where $r_1,r_2,r_3$ are uniformly distributed on $[0,1]$.

Otherwise said, as a two-step barycentration :

$$M=(1-\sqrt{r_{1}})A+\sqrt{r_{1}}\color{red}{\left[r_{2}B+(1-r_{2})C\right]}\tag{2}$$

The earliest version I have found of formulas (1)/(2) is in this interesting ACM 2002 article : see its p. 814 with a graphical "heuristic proof" of formula (2) above given p. 815 that I reproduce here :

Formulas (1)/(2) are mentionned and established here.

Numerous other references and links like this one.

Remark :

My recent encounter with this issue is in this question. My answer attests that I wasn't convinced at first look (maybe because they had been found by Chat GPT) : I was totally wrong.

Answer 3

You could sample uniformly a point in the square with vertices $(0,0)$, $(1,0)$, $(1,1)$, by taking $(x,y) \in [0,1]^2$ uniformly and then map it to $(\max(x,y), \min(x,y))$. Once you can sample uniformly in one triangle, it can be done in any triangle.

Similarly one can sample uniformly in an $n$-dimensional simplex, for small $n$. Not clear how expensive it would be to rearrange ( or test the order) if $n$ is say $10000$.