Generate random locations within a triangular domain

Question

I want to generate x and y having a uniform distribution and limited by [xmin,xmax] and [ymin,ymax]

The points (x,y) should be inside a triangle.

How can I solve such a problem?

Cheerly

The requirements here seem underspecified (and possibly contradictory): if x and y are independent and each is uniformly distributed on an interval, they'll cover a rectangle rather than a triangle. How is this triangle specified, and what's its relation to the range [xmin, xmax] and [ymin, ymax]? – Haberdashery

Don’t just tell us what you want (especially underspecified). What have you tried? What’s gone wrong? Where are you stuck? – Duprey

We probably all guessed that min, max relates to the triangle coordinates. But looking at her other question, I am afraid, we won't get any feedback. – Xerosis

See mathworld.wolfram.com/TrianglePointPicking.html – Throughout

Answer 1

Here's some code that generates points uniformly on an arbitrary triangle in the plane.

import random
    
def point_on_triangle(pt1, pt2, pt3):
    """
    Random point on the triangle with vertices pt1, pt2 and pt3.
    """
    x, y = sorted([random.random(), random.random()])
    s, t, u = x, y - x, 1 - y
    return (s * pt1[0] + t * pt2[0] + u * pt3[0],
            s * pt1[1] + t * pt2[1] + u * pt3[1])

The idea is to compute a weighted average of the three vertices, with the weights given by a random break of the unit interval [0, 1] into three pieces (uniformly over all such breaks). Here x and y represent the places at which we break the unit interval, and s, t and u are the length of the pieces following that break. We then use s, t and u as the barycentric coordinates of the point in the triangle.

Here's a variant of the above that avoids the need to sort, instead making use of an absolute value call:

def point_on_triangle2(pt1, pt2, pt3):
    """
    Random point on the triangle with vertices pt1, pt2 and pt3.
    """
    x, y = random.random(), random.random()
    q = abs(x - y)
    s, t, u = q, 0.5 * (x + y - q), 1 - 0.5 * (q + x + y)
    return (
        s * pt1[0] + t * pt2[0] + u * pt3[0],
        s * pt1[1] + t * pt2[1] + u * pt3[1],
    )

Here's an example usage that generates 10000 points in a triangle:

pt1 = (1, 1)
pt2 = (2, 4)
pt3 = (5, 2)
points = [point_on_triangle(pt1, pt2, pt3) for _ in range(10000)]

And a plot obtained from the above, demonstrating the uniformity. The plot was generated by this code:

import matplotlib.pyplot as plt
x, y = zip(*points)
plt.scatter(x, y, s=0.1)
plt.show()

Here's the image:

And since you tagged the question with the "numpy" tag, here's a NumPy version that generates multiple samples at once. Note that it uses the matrix multiplication operator @, introduced in Python 3.5 and supported in NumPy >= 1.10. You'll need to replace that with a call to np.dot on older Python or NumPy versions.

import numpy as np

def points_on_triangle(v, n):
    """
    Give n random points uniformly on a triangle.

    The vertices of the triangle are given by the shape
    (2, 3) array *v*: one vertex per row.
    """
    x = np.sort(np.random.rand(2, n), axis=0)
    return np.column_stack([x[0], x[1]-x[0], 1.0-x[1]]) @ v


# Example usage
v = np.array([(1, 1), (2, 4), (5, 2)])
points = points_on_triangle(v, 10000)

Haberdashery

Your answer was instrumental in me writing a quick Sierpinski Triangle generator, thought I'd share. – Rompers

u could simply be 1-s-t. – Jammiejammin

Answer 2

Ok, time to add another version, I guess. There is known algorithm to sample uniformly in triangle, see the paper by Osada et al. (2002), chapter 4.2 for details.

Python code:

import math
import random

import matplotlib.pyplot as plt

def trisample(A, B, C):
    """
    Given three vertices A, B, C, 
    sample point uniformly in the triangle
    """
    r1 = random.random()
    r2 = random.random()

    s1 = math.sqrt(r1)

    x = A[0] * (1.0 - s1) + B[0] * (1.0 - r2) * s1 + C[0] * r2 * s1
    y = A[1] * (1.0 - s1) + B[1] * (1.0 - r2) * s1 + C[1] * r2 * s1

    return (x, y)

random.seed(312345)
A = (1, 1)
B = (2, 4)
C = (5, 2)
points = [trisample(A, B, C) for _ in range(10000)]

xx, yy = zip(*points)
plt.scatter(xx, yy, s=0.2)
plt.show()

And result looks like

Osada, R., Funkhouser, T., Chazelle, B., & Dobkin, D. (2002). Shape distributions. ACM Transactions on Graphics (TOG), 21(4), 807-832.

Arabel

Answer 3

Uniform on the triangle?

import numpy as np

N = 10 # number of points to create in one go

rvs = np.random.random((N, 2)) # uniform on the unit square
# Now use the fact that the unit square is tiled by the two triangles
# 0 <= y <= x <= 1 and 0 <= x < y <= 1
# which are mapped onto each other (except for the diagonal which has
# probability 0) by swapping x and y.
# We use this map to send all points of the square to the same of the
# two triangles. Because the map preserves areas this will yield 
# uniformly distributed points.
rvs = np.where(rvs[:, 0, None]>rvs[:, 1, None], rvs, rvs[:, ::-1])

Finally, transform the coordinates
xmin, ymin, xmax, ymax = -0.1, 1.1, 2.0, 3.3
rvs = np.array((ymin, xmin)) + rvs*(ymax-ymin, xmax-xmin)

Uniform marginals? The simplest solution would be to uniformly concentrate the mass on the line (ymin, xmin) - (ymax, xmax)

rvs = np.random.random((N,))
rvs = np.c_[ymin + (ymax-ymin)*rvs, xmin + (xmax-xmin)*rvs]

but that is not very interesting, is it?

Garland