In the original problem statement, you could solve by re-assigning your green region to be orange, your red region to be blue, your black region to green, and the white regions to red. Also note that regions that meet at a single point (as in your example) are not considered to be adjacent.
There is no limit to how many regions can surround and be in contact with a given region, it's just that the shared border that defines "in contact with" must be of non-zero length.
Note also: we are dealing with finite maps. That means that for any given map there will be a shortest boundary for that map, but it doesn't mean there is a minimum constraint.
Consider the gray pixel in the center a placeholder, much like Escher's place holder in the above image (because he couldn't think of how to realistically depict what turns out to be the infinite fractal nature of the center)
> An edge, no matter how small breaks the cross.
My very weak intuition says that an edge of infinite smallness would not.
Now make N as large as you like.
So allowing this makes the problem uninteresting, and precluding it makes the problem interesting and hard.
(I don't recall details - but the preconditions of the Four Colour Theorem rule out all the fuzzy sets, infinitely crooked lines, "this region is all points with one rational and one irrational coordinate", and other trickery that folks who've had a bit of math might otherwise be tempted by.)
I have always wondered about this sample. Consider the possible 4-colours to be red, blue, green and orange, as in the image. We need to determine the colour of the black square. All 4 colours are joined in this inverted cross like formation and the center of the cross (a gray pixel atm) kind of indicates an infinity point. If the space in which these colours are drawn is not discrete, then they can all be said to join at the center of the cross.
Is this sample just not allowed by the rules of the "game"?
Consider it an extreme case of gerry-mandering...