"quasi" <email@example.com> wrote in message news:firstname.lastname@example.org... > Mike Terry wrote: > >quasi wrote in message > >>David Hartley wrote: > >>>quasi wrote: > >>>> > >>>>(2) If S is a closed (but not necessarily simple) polygonal > >>>>curve, then V = R^2. > >>>> > >>>>James Waldby's very simple counterexample disproves (1). > >>>> > >>>>But (2) is still alive, and I'm convinced it's true. It's sort > >>>>of "obviously true", but I've been fooled before by flawed > >>>>visualizations, so even if it really is obviously true, that's > >>>>not good enough -- proof is required. > >>> > >>>Just take scattered's example, with the lines slightly > >>>separated, and then join each extended line back to the end of > >>>the line it was originally joined too. As long as the > >>>extensions are long enough the new joining pieces are not > >>>"visible" from the centre. > >> > >>Nice. > >> > >>Ok, then I'll retreat to this: > >> > >>(2') If S is a simple closed polygonal curve (with finitely > >>many edges), then V contains all points of R^2 which are > >>outside of S. > > > >This still doesn't work - we can make a saw-tooth shape, >"surrounding" a central point in the sense that all the point > >can "see" is saw teeth, each of which is partly obscured by > >another tooth. The teeth are mostly directly joined up to > >their adjacent teeth, but the last two of them join by a > >polygonal curve going right around the outside of all the > >teeth. This puts our central point on the outside of the > >polygon... > > I tried, but I just can't picture it (ny visualization skills > are not very strong).
hmm, ok I'll try an ASCII diagram...
Start with 4 lines: G | | | A-------------------H | | B | | | | | | O | | | | | | F | | D---------------------E | | | C
O is our observation point. AH is partially obscured by BC. BC is partially obscured by DE. DE is partially obscured by GF. GF is partially obscured by AH.
We now join them up, but note that no new lines become visible from O in the process: Join A to B directly Join C to D directly Join E to F directly Join G to H around the outside, i.e. From G we head off to the right, then down past E, then left past C, up past A, and finally joining to H.
So O is outside the polygon, but can't see any complete segment.