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While reading about the four colour theorem, I've found some information that for many simple maps, three colours are in fact enough to provide the colouring. What constitues a "simpler" map, though?

From what I've been able to google on the subject, it seems to be NP-complete to decide whether a given map can be three-coloured but maybe there are at least some giveaways about whether a given map is most likely possible to be three-coloured? Or some telltale signs that, while do not work for each and every map, provide a good approximation that a given map can, in fact, be three-coloured?

Straightfw
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    See this question https://math.stackexchange.com/questions/459773/get-the-adjacency-matrix-of-the-dual-of-a-3-connected-k-regular-g-without-pe ... draks is interested in trivalent planar (3-connected) graphs whose faces are only squares & hexagons ... the first few of these only require $3$ colours ! – Donald Splutterwit Apr 28 '17 at 20:07

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It is indeed NP-complete to decide if a planar graph is $3$-colourable (see the paper "Some simplified NP-complete graph problems"). So it is unlikely we will ever see a nice characterization of planar $3$-colourable graphs. There are some known results however that give classes of planar $3$-colourable graphs.

Outerplanar graphs (planar graphs where all vertices lie on the outerface) are known to be $3$-colourable.

Series Parallel graphs (graphs not containing a $K_4$-minor) are known to be $3$-colourable. This is Hadwinger's conjecture (the $4$-colour theorem on steroids) applied to $K_{4}$-minors.

Grotzech's theorem says that planar triangle free graphs are $3$-colourable. In general, if a graph contains no cycles of length $0 \bmod 3$ then the graph is $3$-colourable. There are a bunch of other generalizations of this result to get slightly wider sets of (planar) graphs.

Brook's theorem says that all sub-cubic (max degree $3$) graphs are $3$-colourable.

Graphs with degeneracy $2$ (all subgraphs have a vertex of degree $2$) are $3$-colourable.

Of course, if a graph has no odd cycles, then it is $2$-colourable, so bipartite planar graphs are $2$-colourable.

Ben
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