Is it true that you need atleast 350 colors to color-in the boundries of a flat map where all the boundries are clearly defined?

Question

Since a providence may build on a bridge, and since certain bridges may be placed as archways over land: it is an obvious fact that this Earth could be turned into a topological and territoried donut/torroid; therefore, l assume it is one, needing 7 colors to be properly colored-in as a globe. As a flat map however; the underside and underneath portions of the bridge [and their territories] would be invisible, unless we gave them something akin to transparency; so let's do that. We will add a color for every possible permutation of overlap between the seven types of colorful territory. A bridge has: a "top" (where people walk); an "underside" (that people look up at, and can hang from); and an "underneath" (which is a path that passes under, both: the top, and the underside). This means that there are atleast 3 positions that a color can be on, so we will give this model the variable: "K" = 3; and the variable "N" = 7. To get every permutation of every ordered color-overlap which can appear on the bridge, we simple do: 7^3, to get the answer: 343. Then we add: 7 to 343, because there is also the possibility of surrounding flat areas, which gives us the answer 350. Right? So does one need atleast 350 colors to adequately color-in the boundries of a map of any given real places on the surface our Earth? Am I being too generous? Does it need more?

Answer 1

This is not true. You can color any map on a torus of genus 1 (that is a donut, or a normal cup with one handle) with at most 7 colors. But the higher the genus, the more colors you need. Any bridge you add increases the genus, so there is no a-priori limit on the number of colors needed.

There is a formula giving a lower bound to the minimum number of colors for any map on an orientable surface (like Earth with $n$ bridges), which has been conjectured in 1890 and completely proven in 1968. This is a lower bound, which gives $4$ for a sphere. It doesn't say that the bound is strict, there might be surfaces where you can't attain the bound. Still the bound is increasing and unlimited, the higher the genus (number of bridges) the more colors you need.

This is the formula: $$ \left\lfloor \frac{7+\sqrt{1+48g}}{2} \right\rfloor $$

Answer 2

I don't understand where the number $7$ is coming from. If you allow the bridges, then we are not colouring planar graphs anymore: we are colouring arbitrary graphs and, for that, no finite number of colours would suffice.

Imagine you have $n$ countries, and imagine there is a bridge between each two of them. (The bridges may be very long and very high, some bridges may need to go over other bridges.)

Think: how many colours do you need in that case? $n$, I guess, as a minimum, just to colour the territories, plus whatever else you want to add in order to colour the rest.