EventHorizon

Yeah, projections aren't the same as slices as they include the interior (guess I should have thought about it more). With point pairs it makes sense how it would work.

This would require connecting 3 1D circles, which would be line segments if you think of the pattern of removing slices or using shadows/projections from higher order circles to lower the dimensionality. If you have two line segments that touch at 1 point on a line, the only line segment on that line that touches both and doesn't intersect either is the line that starts and ends at that point the other two lines share (eg, [-1,0], [0,0], and [0,1]). I don't think this necessarily works due to the third "line" degenerating to a 0D point, but it's the only solution I see that follows the equation, has non-overlapping and unique line segments, and stays in the right number of dimensions.

After a little searching online I found a better solution...

I'll work on "not easy" later.

If you have two tangent circles on a plane and rotate one along the axis going through the centers of both circles some arbitrary angle (so it is now only intersects the plane at two points), would they still be considered tangent? My proof assumes that to be tangent two circles must have the same tangent line at the point where they touch (regardless of which dimensions they pass through). If they just touch (approaching the point from different angles) I don't believe those are tangent. I would change the first part of this sentence in the proof though... "[For this line to be tangent with both existing circles], it would need to be the tangent line between the two existing circles." to [since a circle on this plane could only be tangent to space S at one point and in the direction of this line] But where any two circles do touch, they have the same tangent line at that point. The restriction to a line was because that was the only place that the plane containing the new circle would intersect the current 3d space (and it could have been explained better... fixed above though). If this definition/restriction on tangency in 3d does not work for you, there's a simple example of an infinite number of circles being "tangent" to each other. In fact, they are all "tangent" to each other at two points each. Simply choose unique circles (avoiding the tangency points of the circles already chosen) of radius r from a sphere of radius r.

High dimensional geometry isn't quite in my field of expertise, so if someone sees an error in my proof let me know.

Question (and a hint) for slyde87 and peanut.

That's a nice variable-time solution, but I am looking for exactly equal... or a proof that no such solution exists. Also, infinity is not acceptable (or this question would be no different that the first).

As for your answers to the original problem, I believe they both will work. As for your solution for a specified number of coin flips: it was a good attempt, but...

Sorry for resurrecting an old post, but I found some relevant math. First, here are the links to where I found the information. Tetration Lambert W function The inverse of the function y = x^x is x = (ln y) / W(ln y), where ln is the natural log (aka base e) and W is the Lambert W function. The value of the Lambert W function can be approximated using Newton's method as follows (where wi is the ith appoximation in the series whose limit is the sought value). (Taken directly from wikipedia, w=W(z) so z = we^w) wi+1 = wi - ( wiewi - z ) / ( wiewi + ewi ) The actual limit of xxxxx... = W(-ln x) / (-ln x)

Interesting approach, but it doesn't choose with equal probability.

Exactly what I was thinking, but here's a follow-up question (which isn't that much harder than the original). Is there a solution that is guaranteed to pick an option out of 7 with equal probability within a specified number of uses of the random number generator? If not, prove it.