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).
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ccppurcell
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
> 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.)
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polishdude20
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.
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).
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`.
> 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.)
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.
Interesting and fun