1) a single cell is topologically equivalent to a circle
2) i will treat the idea of a planar surface as intuitive and not explicit, the explicit case of convex folding into the 3rd dimension is included in
mutual adjacency is what requires differentialble colors
1) unbounded/infinite single cell - one color
2) single cell floating in another cell - two color
3)single cell with two mutually adjacent borders [think of a split plane with a circle floating between the two] requires 3 colors
4)single cell with three mutally adjacent borders [think of a plane split with 3 radial (fat pie slice) regions and a circle that is ontop of the
radius point] requires 4 colors
radially this is the maximum number of mutually adjacent cells, if you have 4 radial cells the pairs across from one another are not adjacent. As you
increase the number of radial regions/cells the mutual adjacency of all regions is reduced and never increased.
We want to find the case of maximum adjacency on a planar type surface.
imagine a triangle created by 3 adjacent border colors.
wrap this around a sphere (which creates maximum adjacency with convergence on far side of sphere) which topologically equals a 4 color
we attempt to add a new cell adjacent to all four cells, retaining a convex (sphere-tetrahedron) surface:
there is no way of coloring along the edges of the tetrahedron (attempting to create a new cell) that contacts all the the triangles without isolating
one of them from another meaning they are no longer adjacent and therefore no longer required to be different colors.
you can't color down the middle of a triangle [biforcate it] because that creates two cells that are no longer forced to be the same color. (they
each become cells isolated from each other)
explict proof you cannot biforcate a triangular region where each of the created regions are adjacent to all three border regions
note: biforcating a triangle with a region is equivalent to biforcating it with a flexible line segment
biforcating a triangle:
topological possibilities, biforcating seg can contact triangle either in mid-segment or at point
think of a flexible wire,
wire endpoints contact the border triangle at 2 places:
1. point - itself (same as a cell floating within a cell)
2. point - adjacent midpoint
3. point - another point
4. point - opposite midpoint
5. midpoint - itself (equivilent to case 1, the cell floating within a cell)
6. midpoint - another midpoint
case for each above
(reg=biforcation created region; assume reg1 and reg2 always have contact; border=bordercolor)
1. reg1 contacts no border, reg2 contacts 3 borders (reg1-anything but 4th color, reg2 must be 4th color)
2. reg1 contacts one border, reg2 contacts 3 borders (reg1 non-contact border, reg2 4th color)
3. reg1 contacts one border, reg2 contacts 2 borders (reg1 non-contact border or 4th, reg2 non-contact border or 4th color)
4. reg1 contacts 2 border, reg2 contacts 2 border (reg1 non-contact border or 4th, reg2 non-contact or 4th)
5. (see case 1)
6. reg1 contacts 2 border, reg2 contacts 3 border (reg1 non-contact border, reg2 4th color)
this demonstrates that there is no way of biforcating a triangle where both regions contact all three borders. worst case is 6 with them contacting 2
and 3 border regions.