tag:blogger.com,1999:blog-3829599.post114833869680760959..comments2019-06-06T01:28:31.309-07:00Comments on Agoraphilia: Territorial Teaser: Too Many AnswersUnknownnoreply@blogger.comBlogger5125tag:blogger.com,1999:blog-3829599.post-1148346852632784462006-05-22T18:14:00.000-07:002006-05-22T18:14:00.000-07:00Yeah, that's obviously correct. I guess I didn't t...Yeah, that's obviously correct. I guess I didn't think that through very well. I suppose I should have taken a closer look at the earlier post. The first picture satisfied the criteria I had in mind as long as you count points as intersections. The post that set up the problem was perfectly clear on that. Sorry.FXKLMhttps://www.blogger.com/profile/09870347924405175824noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1148345158160418972006-05-22T17:45:00.000-07:002006-05-22T17:45:00.000-07:00FXKLM --You scared me!When Glen says "a set must b...FXKLM --<BR/><BR/>You scared me!<BR/><BR/>When Glen says "a set must be contiguous", he means that there is a path from any one member of the set to any other member of the set that passes only through faces (not vertices) and does not pass through any member not in the set. True, the two faces you point out do not touch, but the should-be-contiguous set to which they belong, namely, the set of faces with capital C, is in fact contiguous.Jeff Brownhttps://www.blogger.com/profile/10798023289599415591noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1148344227575626492006-05-22T17:30:00.000-07:002006-05-22T17:30:00.000-07:00The solution in the second and third pictures don'...The solution in the second and third pictures don't work. aBC, for instance, is required to touch AbC, which it does not. The simple description of the problem is that every territory is required to touch every other territory with at most one exception. It's easy to see if a proposed solution fails to satisfy that criterion.FXKLMhttps://www.blogger.com/profile/09870347924405175824noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1148343556859602962006-05-22T17:19:00.000-07:002006-05-22T17:19:00.000-07:00(After Ben's colored solution was added to the pos...(After Ben's colored solution was added to the post, I had to update my comment.)<BR/><BR/>For the octahedral characterization to be complete, one would have to allow deformation of intersections into edges – more precisely, declare any vertex where four edges meet equivalent to two three-edge vertices with a common edge. Deforming a four-way intersection into an edge in this way adds another connection between the faces and destroys none, so splitting the vertices of one solution yields another solution. Armed with this equivalence we can generate Ben's colored solution, which is otherwise not strictly topologically equivalent to the octahedron because some of the faces he drew have four neighbors.<BR/><BR/>So far every solution posted fits this characterization, but I still don't know if it's complete.Jeff Brownhttps://www.blogger.com/profile/10798023289599415591noreply@blogger.comtag:blogger.com,1999:blog-3829599.post-1148341450430148382006-05-22T16:44:00.000-07:002006-05-22T16:44:00.000-07:00You can characterize these solutions as “deformati...You can characterize these solutions as “deformations of an octahedron” ... but I don’t know how to prove that that’s a complete characterization of the solution set.<BR/><BR/>The following observation was fun: Imagine slicing a beach ball along three perpendicular axes. Depending on how you projected the resulting figure onto the plane, you could get any of the first four pictures fairly naturally. One controlling parameter is whether you assign the center of the projection to a face (pictures 1 and 4) or a corner (2 and 3). Another parameter is whether you assign the point at infinity to a face (picture 3) or omit it (2).<BR/><BR/>If you projected the beach ball from a face, you get picture 1 (Glen's original). If you "sharpen" the beach ball into an octahedron before projection, you get 4.<BR/><BR/>Conveniently, the beach ball/octahedron solution could be characterized as "overlapping circles".Jeff Brownhttps://www.blogger.com/profile/10798023289599415591noreply@blogger.com