Abstract
Colour the edges of the complete graph with vertex set {1,2,⋯,n}\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$${\\{1, 2, \\dotsc , n\\}}$$\\end{document} with an arbitrary number of colours. What is the smallest integer f(l, k) such that if n>f(l,k)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$n > f(l,k)$$\\end{document} then there must exist a monotone monochromatic path of length l or a monotone rainbow path of length k? Lefmann, Rödl, and Thomas conjectured in 1992 that f(l,k)=lk-1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$f(l, k) = l^{k - 1}$$\\end{document} and proved this for l⩾(3k)2k\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$l \\geqslant (3 k)^{2 k}$$\\end{document}. We prove the conjecture for l⩾k3(logk)1+o(1)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$l \\geqslant k^3 (\\log k)^{1 + o(1)}$$\\end{document} and establish the general upper bound f(l,k)⩽k(logk)1+o(1)·lk-1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$f(l, k) \\leqslant k (\\log k)^{1 + o(1)} \\cdot l^{k - 1}$$\\end{document}. This reduces the gap between the best lower and upper bounds from exponential to polynomial in k. We also generalise some of these results to the tournament setting.
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