Abstract

Let RR and BB be two disjoint sets of points in the plane such that ∣B∣⩽∣R∣|B|\leqslant |R|, and no three points of R∪BR\cup B are collinear. We show that the geometric complete bipartite graph K(R,B)K(R,B) contains a non-crossing spanning tree whose maximum degree is at most max⁡{3,⌈∣R∣−1∣B∣⌉+1}\max\left\{3, \left\lceil \frac{|R|-1}{|B|}\right\rceil + 1\right\}; this is the best possible upper bound on the maximum degree. This solves an open problem posed by Abellanas et al. at the Graph Drawing Symposium, 1996.

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