In this paper we consider the following problem: Given a Hamiltonian graph G, and a Hamiltonian cycle C of G, can we compute a second Hamiltonian cycle C′≠C\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$C^{\\prime } \ e C$$\\end{document} of G, and if yes, how quickly? If the input graph G satisfies certain conditions (e.g. if every vertex of G is odd, or if the minimum degree is large enough), it is known that such a second Hamiltonian cycle always exists. Despite substantial efforts, no subexponential-time algorithm is known for this problem. In this paper we relax the problem of computing a second Hamiltonian cycle in two ways. First, we consider approximating the length of a second longest cycle on n-vertex graphs with minimum degree δ\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\delta $$\\end{document} and maximum degree Δ\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\Delta $$\\end{document}. We provide a linear-time algorithm for computing a cycle C′≠C\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$C^{\\prime } \ e C$$\\end{document} of length at least n-4α(n+2α)+8\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$n-4\\alpha (\\sqrt{n}+2\\alpha )+8$$\\end{document}, where α=Δ-2δ-2\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\alpha = \\frac{\\Delta -2}{\\delta -2}$$\\end{document}. This results provides a constructive proof of a recent result by Girão, Kittipassorn, and Narayanan in the regime of Δδ=o(n)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\frac{\\Delta }{\\delta } = o(\\sqrt{n})$$\\end{document}. Our second relaxation of the problem is probabilistic. We propose a randomized algorithm which computes a second Hamiltonian cycle with high probability, given that the input graph G has a large enough minimum degree. More specifically, we prove that for every 0<p≤0.02\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$0<p\\le 0.02$$\\end{document}, if the minimum degree of G is at least 8plog8n+4\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\frac{8}{p} \\log \\sqrt{8}n + 4$$\\end{document}, then a second Hamiltonian cycle can be computed with probability at least 1-1n50p4+1\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$1 - \\frac{1}{n}\\left( \\frac{50}{p^4} + 1 \\right) $$\\end{document} in poly(n)·24pn\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$poly(n) \\cdot 2^{4pn}$$\\end{document} time. This result implies that, when the minimum degree δ\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\delta $$\\end{document} is sufficiently large, we can compute with high probability a second Hamiltonian cycle faster than any known deterministic algorithm. In particular, when δ=ω(logn)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$\\delta = \\omega (\\log n)$$\\end{document}, our probabilistic algorithm works in 2o(n)\\documentclass[12pt]{minimal} \\usepackage{amsmath} \\usepackage{wasysym} \\usepackage{amsfonts} \\usepackage{amssymb} \\usepackage{amsbsy} \\usepackage{mathrsfs} \\usepackage{upgreek} \\setlength{\\oddsidemargin}{-69pt} \\begin{document}$$2^{o(n)}$$\\end{document} time.