[HN Gopher] Points on a ring: An interactive walkthrough of a po...
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Points on a ring: An interactive walkthrough of a popular math
problem
Author : evakhoury
Score : 43 points
Date : 2026-03-02 17:11 UTC (1 days ago)
(HTM) web link (growingswe.com)
(TXT) w3m dump (growingswe.com)
| atnnn wrote:
| I find this easier to visualize with an equivalent way to pick
| points: drop four (n) lines through the center of the circle. For
| each line, randomly pick one of the two points that intersect the
| circle.
|
| Independent of the lines dropped, there are eight (2n) ways to
| pick adjacent points and sixteen (n^2) different combinations of
| points.
|
| There are four (n) adjacent points iif the points lie on the same
| half of the circle (proof by interactive visualization).
|
| So the answer is eight sixteenths (2n/2^n).
| matheist wrote:
| Great argument. I found this generalization to higher
| dimension: https://www.mathpages.com/home/kmath327/kmath327.htm
| nsvd2 wrote:
| Interesting and fun
| polishdude20 wrote:
| I think about this by unwrapping the circle to form a straight
| line. Then you draw an imaginary point in the middle of the line.
| Then what are the chances they will all fall on one side of the
| line or the other? 1/2 because it's divided into two equal
| lengths.
| mobeets wrote:
| This approach implies the probability doesn't depend on N. It
| only happens to be 1/2 for N=4 (the article goes into this).
| The trick is that you don't know beforehand which semicircle
| all the points can land in, but your unwrapping step assumes
| you do.
| thaumasiotes wrote:
| There are two major problems with your idea:
|
| 1. Your answer can be "no" when the true answer is "yes".
| Consider this process with a circle of perimeter "21":
| --------------------- (unwrap the circle)
| ----------+---------- (bisect the line)
| -**-------+--------** (drop four points)
|
| The four points don't fall into either of the two semicircles
| that you stupidly predefined, but they do fall into a different
| semicircle.
|
| 2. Your answer of "1/2, because it's divided into two equal
| lengths" is completely wrong _for the scenario that you
| specify_.
|
| Consider the case where we drop a single point. We can do the
| same procedure:
|
| A. Unwrap the circle;
|
| B. Bisect the line;
|
| C. Drop one point.
|
| But even though the line is still divided into two equal
| lengths, our one point has a 100% chance of falling either on
| one side of the bisection point, or on the other side.
|
| For the case where we drop four points, the article already
| gives the correct answer for your method, which is 1/2^3
| (because there are 3+1 points).
| ccppurcell wrote:
| Probably best to avoid the word ring in a mathematics discussion
| unless you're talking about the algebraic structure. It's very
| much a mathematical `keyword`.
| matheist wrote:
| > _The same decomposition works in higher dimensions._
|
| I don't think the same argument works in higher dimensions. On a
| circle, we can canonically pick a semicircle corresponding to
| each point (we have two choices, let's say we pick the clockwise
| one).
|
| In higher dimensions there's no canonical choice of half-sphere.
| In odd dimensions one _could_ pick a canonical half-sphere per
| point but it might turn out that some other non-chosen half-
| sphere for that point contains all the other points. In even
| dimensions there isn 't even a way to canonically pick a half-
| sphere for each point (this is a consequence of the Hairy Ball
| Theorem).
|
| (For all I know the actual numbers might turn out to be the same,
| I don't know. I'm just saying that the argument doesn't work.)
| thaumasiotes wrote:
| Well, it's easy to pick a half-n-sphere corresponding to a
| point on the surface. You just take the half-sphere centered at
| that point. For our two-dimensional circle, it would be the arc
| defined by the diameter that is perpendicular to the radius
| running between "the point" and the center of the circle.
|
| At that point you've lost the ability to say that only one such
| half-sphere defined by a dropped point can be a valid solution,
| and you've also lost the ability to say that if a valid
| solution exists then there must be a valid solution defined by
| one of the points you want to include in the half-sphere, but
| you _can_ define a canonical half-sphere for any point.
|
| I was uncomfortable with the idea of picking "random points on
| a circle" to begin with, because of
| https://en.wikipedia.org/wiki/Bertrand_paradox_(probability) ,
| but the article doesn't even address whether the concept is
| well-defined. We can always choose a point on the perimeter
| deterministically from any chord (...that isn't a diameter), so
| the ill-definedness of the problem of choosing a random chord
| seems like it would infect the problem of choosing a random
| point on the perimeter.
| matheist wrote:
| Right, I was taking it as given that the problem of choosing
| a hemisphere canonically for a point meant "such that the
| argument works in the same way as for the circle".
|
| Bertrand paradox just doesn't apply here, there's a natural
| measure on the circle and all higher dimensional spheres. I
| wouldn't expect an article on this subject to need to make
| that clarification unless it's dealing with chords or some
| other situation without a natural measure.
| thaumasiotes wrote:
| If I choose my four points as the endpoints of two chords
| chosen by the "random radial point" method described on the
| wikipedia page, is it still true that the odds of all four
| being covered by a semicircle are 50%?
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(page generated 2026-03-03 23:01 UTC)