Oof, right you are! So my "I've never seen this before" was inadvertently a lie. And now I also want to know about the random constellation answer in dimensions >2! In dim=2 (as in the MSRI puzzle Dan asked), this turns into a nicely discrete problem. If you label the arcs between successive people in the circle 1,2,...n from smallest to largest, then a pair of people looking at each other just corresponds to an integer that is smaller than either of its neighbors. It's not hard to see that all n! permutations are equally likely. I won't give away the expected fraction of numbers are smaller than either of their neighbors, but it's what I found to be a pleasing result. Does this turn into a discrete problem in higher dimensions also, via something like random Voronoi regions on a sphere? --Michael On Wed, Jun 3, 2020 at 11:06 AM James Propp <jamespropp@gmail.com> wrote:
I was never able to figure out the solution to Veit’s puzzle, the first time it came around or now, so at this point I’m prepared to swallow my pride and ask someone to email me the solution. Or maybe someone could send it to the whole list, with a spoiler-alert header page. (I miss ^L from the days of UNIX mail.)
Jim Propp
On Wed, Jun 3, 2020 at 8:09 AM Veit Elser <ve10@cornell.edu> wrote:
On Jun 2, 2020, at 9:50 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Wow, that "expected number of lonely people" problem is great! I've never seen this before, and did not expect it to come out so nicely.
—Michael
That problem can be recast as the “expected constellation size” problem I posed many years ago. It can be worked out in any number of dimensions.
In the night sky (2D) assume that the visible stars are uniformly distributed and form connected graphs (constellations) by joining each
star
with its nearest (angular) neighbor. What’s the expected number of stars per constellation?
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