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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def convex_hull(points):
 """Computes the convex hull of a set of 2D points. Implements `Andrew's monotone chain algorithm <http://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain>`_. The algorithm has O(n log n) complexity. Credit: `<http://en.wikibooks.org/wiki/Algorithm_Implementation/Geometry/Convex_hull/Monotone_chain>`_ Parameters points : list of tuples An iterable sequence of (x, y) pairs representing the points. Returns ------- Output : list A list of vertices of the convex hull in counter-clockwise order, starting from the vertex with the lexicographically smallest coordinates. """


    # Sort the points lexicographically (tuples are compared lexicographically).
    # Remove duplicates to detect the case we have just one unique point.
    points = sorted(set(points))

    # Boring case: no points or a single point, possibly repeated multiple times.
    if len(points) <= 1:
        return points

    # 2D cross product of OA and OB vectors, i.e. z-component of their 3D cross product.
    # Returns a positive value, if OAB makes a counter-clockwise turn,
    # negative for clockwise turn, and zero if the points are collinear.
    def cross(o, a, b):
        return (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])

    # Build lower hull
    lower = []
    for p in points:
        while len(lower) >= 2 and cross(lower[-2], lower[-1], p) <= 0:
            lower.pop()
        lower.append(p)

    # Build upper hull
    upper = []
    for p in reversed(points):
        while len(upper) >= 2 and cross(upper[-2], upper[-1], p) <= 0:
            upper.pop()
        upper.append(p)

    # Concatenation of the lower and upper hulls gives the convex hull.
    # Last point of each list is omitted because it is repeated at the beginning of the other list.
    return lower[:-1] + upper