Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Post History

#1: Initial revision by user avatar Karl Knechtel‭ · 2023-08-26T14:59:15Z (over 1 year ago)
How can I choose a point from a uniform distribution within a regular polygon?
Suppose I have a regular polygon in the Cartesian plane, centered at the origin with circumradius $1$, aligned with one vertex at $(1, 0)$. Given a source of independent, uniformly distributed random variables on $[0..1)$, how can I choose a point within the polygon, randomly, with uniform distribution, using only finitely many inputs?

Is this possible for all regular polygons or only for certain numbers of sides? (Of course the square case is trivial.)

I cannot use [rejection sampling](https://en.wikipedia.org/wiki/Rejection_sampling) within a bounding area (e.g. the unit square) as this requires an unbounded (heh) number of random selections.

I know that a random point can be chosen inside a circle by randomly choosing an angle (mapping one input to $[0..2\pi)$) and distance from the origin (taking the square root of the input variable to correct for the bias). However, this doesn't seem to work for polygons; if I scale the chosen "radius" to fit within the polygon, the result will then show a bias towards the perpendicular bisectors of the sides (and away from the lines connecting the centre to the edges - for the same reason that the square-root correction was necessary).

----

<small>This question is inspired by a [recent Stack Overflow question](https://stackoverflow.com/questions/76983253). I believe I have an answer, but I would prefer to let others offer an answer first.</small>