Two nits ... (a) There's an edge case in the intersecting-plane test: The two circles could lie in the same plane. In this case, they might intersect (or not), or be tangent, but they can't be linked. This isn't a theory problem, but it might be a programming problem: Computing the intersection of two nearly-parallel planes is a numerical challenge. We could have two circles that have no chance of linkage, because they are moderately far apart, but still cause a "can't tell" in the linked-p subroutine. (b) Fred's separability test seems to work if we allow spheres as separators (instead of just allowing planes). Rich --------- Quoting Fred lunnon <fred.lunnon@gmail.com>:
Dan has very politely pointed out that I've been talking through my hat again.
To start with, there are unlinked cicles in 3-space which are not separated by a plane (prime), which demolishes my proposed "criterion".
Secondly, the line in Dan's criterion is the meet of the planes in which the respective circles lie --- what I had in mind as a decision "algorithm".
Thirdly, in general a k-sphere and l-sphere lie in a (k+1)-flat and (l+1)-flat, which again meet in a line in (k+l+1)-space: there's no need for any induction.
Fred Lunnon
On 10/20/09, Fred lunnon <fred.lunnon@gmail.com> wrote:
On 10/20/09, Dan Asimov <dasimov@earthlink.net> wrote:
...
Yes. In fact, for disjoint round spheres S^p and S^q in R^n where p+q+1 = n, they link exactly when there exists a line L in R^n that intersects each sphere twice in alternating order.
(In other words, L intersects the spheres in two linked 0-spheres.)
--Dan
Completely different from the criterion I had in mind --- that they are not linked just when there exists a prime disjoint from and separating them!
And Dan's criterion does yield an algorithm --- the line in question simply joins their centres. WFL
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