. Consider-planar-k-bmrn-colouring, which is the restriction of k-BMRN-COLOURING to planar spanned digraphs. For k = 2, the problem is polynomial-time solvable because k-BMRN-COLOURING is polynomial-time solvable

, Proof: The proof is by reduction from PLANAR 3-COLOURING which consists in deciding whether a given planar graph is 3-colourable

, From a planar graph G, we shall construct

G. Since and . Planar, Moreover, free to move slightly some vertices, we may assume that no two vertices ofG lie on a same horizontal or vertical line. Let G be the orientation of G obtained by orienting every edge towards its higher vertex inG: hence if uv ? A( G), then vertex v lies above u

, preserving the colouring equivalence with G. FromG, one can easily derive a plane straight-line embedding of D G in which all arcs u x v x , for x ? V ( G), are drawn horizontally and from left to right, of tiny length and whose middle is x. Towards getting (D, T ), we first add

?. , T. )-where-d-=-d-g-?-x?v-(g)-p-x, T. =-m-g-?-x?v-(g)-p-x-;-d-g, and M. Conversely, Of course, we do this in such a way that the added paths do not cross, Let us now check that BMRN(D G , M G ) ? 3 if and only if BMRN(D, T ) ? 3. Since (D G , M G ) is the restriction of (D, T ) to V (D G ), every 3-BMRN-colouring of (D, T ) induces a 3-BMRN-colouring of

. Thus, T ) is a planar spanned digraph that is 3-BMRN-colourable if and only if G is 3-colourable

, PLANAR k-COLOURING is trivial for all k ? 4, as the answer is always 'Yes' according to the FourColour Theorem. Henceforth, the above proof cannot be generalized to show that PLANAR

, the complexity of PLANAR k-BMRN-COLOURING? Similarly, one can compute the chromatic number of an outerplanar graph in polynomial time. So we cannot establish the hardness of the restriction of k-BMRN-COLOURING to outerplanar digraphs via a proof similar to that of Theorem 5.8. We thus leave the following question open: Question 5.10. For any outerplanar spanned digraph

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