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A note on the 2-colored rectilinear crossing number of random point sets in the unit square

Let S\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$S$$\\end{document} be a set of four points chosen independently, uniformly at random from a square. Join every pair of points of S\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$S$$\\end{document} with a straight line segment. Color these edges red if they have positive slope and blue, otherwise. We show that the probability that S\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$S$$\\end{document} defines a pair of crossing edges of the same color is equal to 1/4\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$1/4$$\\end{document}. This is connected to a recent result of Aichholzer et al. [1] who showed that by 2-colouring the edges of a geometric graph and counting monochromatic crossings instead of crossings, the number of crossings can be more than halved. Our result shows that for the described random drawings, there is a coloring of the edges such that the number of monochromatic crossings is in expectation 12-750\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\frac{1}{2}-\\frac{7}{50}$$\\end{document} of the total number of crossings.

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