Abstract
A geometric graph is a graph whose vertices are points in the plane and whose edges are straightline segments between the points. A plane spanning tree in a geometric graph is a spanning tree that is noncrossing. Let R and B be two disjoint sets of points in the plane such that \(R\cup B\) is in general position, and let \(n=R\cup B\). Assume that the points of R are colored red and the points of B are colored blue. A bichromatic plane spanning tree is a plane spanning tree in the complete bipartite geometric graph with bipartition (R, B). In this paper we consider the maximum bichromatic plane spanning tree problem, which is the problem of computing a bichromatic plane spanning tree of maximum total edge length.

1.
For the maximum bichromatic plane spanning tree problem, we present an approximation algorithm with ratio 1 / 4 that runs in \(O(n\log n)\) time.

2.
We also consider the multicolored version of this problem where the input points are colored with \(k>2\) colors. We present an approximation algorithm that computes a plane spanning tree in a complete kpartite geometric graph, and whose ratio is 1 / 6 if \(k=3\), and 1 / 8 if \(k\geqslant 4\).

3.
We also revisit the special case of the problem where \(k=n\), i.e., the problem of computing a maximum plane spanning tree in a complete geometric graph. For this problem, we present an approximation algorithm with ratio 0.503; this is an extension of the algorithm presented by Dumitrescu and Tóth (Discrete Comput Geom 44(4):727–752, 2010) whose ratio is 0.502.

4.
For points that are in convex position, the maximum bichromatic plane spanning tree problem can be solved in \(O(n^3)\) time. We present an \(O(n^5)\)time algorithm that solves this problem for the case where the red points lie on a line and the blue points lie on one side of the line.
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References
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Acknowledgements
We would like to thank an anonymous referee whose comments improved the readability of the paper. Funding was provide by NSERC and NSF (Grant CCF1228639)
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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
A preliminary version of this paper has been accepted to Algorithms and Data Structures Symposium (WADS), 2017.
The work of Ahmad Biniaz has been done while he was at Carleton University.
Appendix: Proof of \(f(x,\alpha ) \geqslant 0\)
Appendix: Proof of \(f(x,\alpha ) \geqslant 0\)
We want to show that
for all \(\sqrt{3}/2\leqslant x \leqslant \sqrt{3}\) and \(0\leqslant \alpha \leqslant \frac{\pi }{6}\). From elementary trigonometry, we have
from which \(f(x,\alpha )\) can be rewritten as
Let us solve the equation \(f(x,\alpha ) = 0\), which corresponds to
By squaring on both sides, we find
which we write as
Squaring once more, we find
which is equivalent to
which can be factored into
Since \(x > 0\) and
we have that \(f(x,\alpha ) = 0\) if and only if \(x = \cos \alpha +\frac{1}{\sqrt{3}}\sin \alpha \). Therefore, on its domain, f is equal to 0 precisely on the curve \(x = \cos \alpha +\frac{1}{\sqrt{3}}\sin \alpha \) and nowhere else. Thus, below this curve, f is everywhere positive or everywhere negative. Since \(f\left( \frac{\sqrt{3}}{2},\frac{\pi }{6}\right) = 1\frac{\sqrt{3}}{2} > 0\), f is everywhere positive below the curve. Similarly, since \(f(\sqrt{3},0) = \sqrt{4\sqrt{3}}  1 > 0\), f is everywhere positive above the curve. Therefore, \(f(x,\alpha ) \geqslant 0\) for all \(\sqrt{3}/2\leqslant x \leqslant \sqrt{3}\) and \(0\leqslant \alpha \leqslant \frac{\pi }{6}\).
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Biniaz, A., Bose, P., Crosbie, K. et al. Maximum Plane Trees in Multipartite Geometric Graphs. Algorithmica 81, 1512–1534 (2019). https://doi.org/10.1007/s004530180482x
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Keywords
 Multipartite geometric graphs
 Plane spanning trees
 Maximum spanning trees
 Approximation algorithms