[seqfan] Re: partition of a circle
njasloane at gmail.com
Sat May 12 06:03:45 CEST 2012
Ed, I disagree! Please read my last message again.
In your example, the left piece and the right piece
have a common boundary which is not connected,
and so this is not admissible.
The pieces, as I said, do NOT need to be assumed to be convex.
On Fri, May 11, 2012 at 11:05 PM, Ed Jeffery <lejeffery7 at gmail.com> wrote:
> Sorry to butt in again, but I think Valette-Z. mean that the pieces have
> null intersection except at their boundaries, with their disjoint union
> being the entire disk (circle); i.e., as with tiles, the pieces must fit
> together without gaps or overlaps.
> As for convexity, surely the pieces must be convex, since otherwise the
> example (for n = 3) I sent to you would be a valid candidate. To describe
> that example again: draw a smaller circle inside the larger circle to be
> divided, then draw two lines, each extending from the smaller circle to the
> larger one, and which divide the annulus into two pieces without either
> line intersecting the interior of the smaller circle or disk. The two
> pieces forming the annulus can't be convex.
> > Wolfdieter, Yes, I agree that we need more conditions
> > that will rule out your example.
> > In fact in the Valette-Z. paper, they add the condition
> > that the partition must be admissible, which
> > they define to mean that if K and L are two distinct pieces,
> > then the intersection (boundary of K) intersect (boundary of L)
> > is connected.
> > This rules out your example, since the two big pieces has intersection
> > which is not connected.
> > By the way, they do NOT require that the pieces be convex. (That would be
> > another way to
> > rule out your example)
> > Neil
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