Algorithm Implementation/Geometry/Convex hull/Monotone chain

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Upper and lowers hulls of a set of points

Andrew's monotone chain convex hull algorithm constructs the convex hull of a set of 2-dimensional points in O(n \log n) time.

It does so by first sorting the points lexicographically (first by x-coordinate, and in case of a tie, by y-coordinate), and then constructing upper and lower hulls of the points in O(n) time.

An upper hull is the part of the convex hull, which is visible from the above. It runs from its rightmost point to the leftmost point in counterclockwise order. Lower hull is the remaining part of the convex hull.

Contents

Pseudo-code [edit]

Input: a list P of points in the plane.

Sort the points of P by x-coordinate (in case of a tie, sort by y-coordinate).

Initialize U and L as empty lists.
The lists will hold the vertices of upper and lower hulls respectively.

for i = 1, 2, ..., n:
    while L contains at least two points and the sequence of last two points
            of L and the point P[i] does not make a counter-clockwise turn:
        remove the last point from L
    append P[i] to L

for i = n, n-1, ..., 1:
    while U contains at least two points and the sequence of last two points
            of U and the point P[i] does not make a counter-clockwise turn:
        remove the last point from U
    append P[i] to U

Remove the last point of each list (it's the same as the first point of the other list).
Concatenate L and U to obtain the convex hull of P.
Points in the result will be listed in counter-clockwise order.

Python [edit]

def convex_hull(points):
    """Computes the convex hull of a set of 2D points.
 
    Input: an iterable sequence of (x, y) pairs representing the points.
    Output: a list of vertices of the convex hull in counter-clockwise order,
      starting from the vertex with the lexicographically smallest coordinates.
    Implements Andrew's monotone chain algorithm. O(n log n) complexity.
    """
 
    # 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[:-1]
 
 
# Example: convex hull of a 10-by-10 grid.
assert convex_hull([(i/10, i%10) for i in range(100)]) == [(0, 0), (9, 0), (9, 9), (0, 9)]

Ruby [edit]

# the python code converted to ruby syntax
def convex_hull(points)
  points.sort!.uniq!
  return points if points.length < 3
  def cross(o, a, b)
    (a[0] - o[0]) * (b[1] - o[1]) - (a[1] - o[1]) * (b[0] - o[0])
  end
  lower = Array.new
  points.each{|p|
    while lower.length > 1 and cross(lower[-2], lower[-1], p) <= 0 do lower.pop end
    lower.push(p)
  }
  upper = Array.new
  points.reverse_each{|p|
    while upper.length > 1 and cross(upper[-2], upper[-1], p) <= 0 do upper.pop end
    upper.push(p)
  }
  return lower[0...-1] + upper[0...-1]
end
fail unless convex_hull((0..9).to_a.repeated_permutation(2).to_a) == [[0, 0], [9, 0], [9, 9], [0, 9]]

C++ [edit]

// Implementation of Andrew's monotone chain 2D convex hull algorithm.
// Asymptotic complexity: O(n log n).
// Practical performance: 0.5-1.0 seconds for n=1000000 on a 1GHz machine.
#include <algorithm>
#include <vector>
using namespace std;
 
typedef int coord_t;         // coordinate type
typedef long long coord2_t;  // must be big enough to hold 2*max(|coordinate|)^2
 
struct Point {
        coord_t x, y;
 
        bool operator <(const Point &p) const {
                return x < p.x || (x == p.x && y < p.y);
        }
};
 
// 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.
coord2_t cross(const Point &O, const Point &A, const Point &B)
{
        return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x);
}
 
// Returns a list of points on the convex hull in counter-clockwise order.
// Note: the last point in the returned list is the same as the first one.
vector<Point> convex_hull(vector<Point> P)
{
        int n = P.size(), k = 0;
        vector<Point> H(2*n);
 
        // Sort points lexicographically
        sort(P.begin(), P.end());
 
        // Build lower hull
        for (int i = 0; i < n; i++) {
                while (k >= 2 && cross(H[k-2], H[k-1], P[i]) <= 0) k--;
                H[k++] = P[i];
        }
 
        // Build upper hull
        for (int i = n-2, t = k+1; i >= 0; i--) {
                while (k >= t && cross(H[k-2], H[k-1], P[i]) <= 0) k--;
                H[k++] = P[i];
        }
 
        H.resize(k-1);
        return H;
}

Perl [edit]

use Sort::Key::Radix qw(nsortkey); # use radix sort for an O(n) algorithm
 
sub convex_hull {
    return @_ if @_ < 2;
 
    my @p = nkeysort { $_->[0] } @_;
 
    my (@u, @l);
    my $i = 0;
    while ($i < @p) {
        my $iu = my $il = $i;
        my ($x, $yu) = @{$p[$i]};
        my $yl = $yu;
        # search for upper and lower Y for the current X
        while (++$i < @p and $p[$i][0] == $x) {
            my $y = $p[$i][1];
            if ($y < $yl) {
                $il = $i;
                $yl = $y;
            }
            elsif ($y > $yu) {
                $iu = $i;
                $yu = $y;
            }
        }
        while (@l >= 2) {
            my ($ox, $oy) = @{$l[-2]};
            last if ($l[-1][1] - $oy) * ($x - $ox) < ($yl - $oy) * ($l[-1][0] - $ox);
            pop @l;
        }
        push @l, $p[$il];
        while (@u >= 2) {
            my ($ox, $oy) = @{$u[-2]};
            last if ($u[-1][1] - $oy) * ($x - $ox) > ($yu - $oy) * ($u[-1][0] - $ox);
            pop @u;
        }
        push @u, $p[$iu];
    }
 
    # remove points from the upper hull extremes when they are already
    # on the lower hull:
    shift @u if $u[0][1] == $l[0][1];
    pop @u if @u and $u[-1][1] == $l[-1][1];
 
    return (@l, reverse @u);
}

C [edit]

C sources taken from the Math::ConvexHull::MonotoneChain Perl module. Note that this implementation works on sorted input points. Otherwise, it's rather similar to the C++ implementation above.

typedef struct {
  double x;
  double y;
} point_t;
 
typedef point_t* point_ptr_t;
 
/* Three points are a counter-clockwise turn if ccw > 0, clockwise if
 * ccw < 0, and collinear if ccw = 0 because ccw is a determinant that
 * gives the signed area of the triangle formed by p1, p2 and p3.
 */
static double
ccw(point_t* p1, point_t* p2, point_t* p3)
{
  return (p2->x - p1->x)*(p3->y - p1->y) - (p2->y - p1->y)*(p3->x - p1->x);
}
 
/* Returns a list of points on the convex hull in counter-clockwise order.
 * Note: the last point in the returned list is the same as the first one.
 */
void
convex_hull(point_t* points, ssize_t npoints, point_ptr_t** out_hull, ssize_t* out_hullsize)
{
  point_ptr_t* hull;
  ssize_t i, t, k = 0;
 
  hull = *out_hull;
 
  /* lower hull */
  for (i = 0; i < npoints; ++i) {
    while (k >= 2 && ccw(hull[k-2], hull[k-1], &points[i]) <= 0) --k;
    hull[k++] = &points[i];
  }
 
  /* upper hull */
  for (i = npoints-2, t = k+1; i >= 0; --i) {
    while (k >= t && ccw(hull[k-2], hull[k-1], &points[i]) <= 0) --k;
    hull[k++] = &points[i];
  }
 
  *out_hull = hull;
  *out_hullsize = k;
}

References [edit]

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