## 2D Convex Hulls: Jarvis March

The simple approach to computing the convex hull in 2D.

I’ve found myself coding convex hull algorithms on a few occasions now, so I decided to implement a few and talk about them here, in case someone new to the subject wants to get the quick ‘n’ dirty. The algorithms I will talk about are the Jarvis March (code), the Graham Scan (code) and Chan’s algorithm (code). I feel they are all different enough that each is worth looking at, but similar enough that they are worth looking at together. I will try and focus more on the actual implementation of the algorithms (all in Python), looking at potential pit falls as well as the niceties, rather then just rehashing what can be found on Wikipedia. This first post looks at the Jarvis March.

### Jarvis March

The Jarvis March is probably the simplest convex hull algorithm conceptually. You start with an extreme point p (a vertex of the convex hull) of a point set. You then find the next point on the convex hull of the point set in CCW order. This is done by finding the “furthest right” point, relative to p. The furthest right point q, relative to p, is the point such all other points in the point set lie to the left of the directed line through p, q (this line lies on the hull boundary). We can then update p to q and repeat the process until we end up with the point we started at.

The heart of the algorithm really lies in finding the furthest right point q, relative to an extreme point p. Say you have 3 points p, q and r. We say p, q, r form a left (respectively right) turn if r lies to the left (right) of the directed line through p and q. We can write a simple function to determine the turn of 3 points:

TURN_LEFT, TURN_RIGHT, TURN_NONE = (1, -1, 0)

def turn(p, q, r):
"""Returns -1, 0, 1 if p,q,r forms a right, straight, or left turn."""
return cmp((q[0] - p[0])*(r[1] - p[1]) - (r[0] - p[0])*(q[1] - p[1]), 0)

Finding the furthest right point relative to p just reduces to simply finding the minimum point q using turn with a fixed parameter p as our comparison function. This can is done in $$O(n)$$ time:

def _next_hull_pt(points, p):
"""Returns the next point on the convex hull in CCW from p."""
q = points[0] != p and points[0] or points[1]
for r in (x for x in points if x != p):
if turn(p, q, r) == TURN_RIGHT:
q = r
return q

The above function is simple, but it assumes the points are in general position; that there are no 3 collinear points. The first problem arises if there are 2 furthest right points, then we could chose the one that is closer to p, which is not an extreme point (ie. it lies on the hull boundary, but is not a vertex). This becomes a bigger problem if our first comparison relative to this “mid-point” is with the 2 vertices on either side of it. We could possibly “skip” over the furthest right point, since it is not to the right of the furthest left, but collinear and would end up with the incorrect furthest right point. Luckily, handling this is rather simple; we insist that if there is more than 1 furthest right point, we choose the one furthest from p. We can then rewrite our function, handling this case:

def _dist(p, q):
"""Returns the squared Euclidean distance between p and q."""
dx, dy = q[0] - p[0], q[1] - p[1]
return dx * dx + dy * dy

def _next_hull_pt(points, p):
"""Returns the next point on the convex hull in CCW from p."""
q = p
for r in points:
t = turn(p, q, r)
if t == TURN_RIGHT or t == TURN_NONE and _dist(p, r) > _dist(p, q):
q = r
return q

The only problem now is finding the first extreme point. This can be done rather simply by choosing the minimum point in lexicographical order. We can then put this together and simply loop until we end up where we started:

def convex_hull(points):
"""Returns the points on the convex hull of points in CCW order."""
hull = [min(points)]
for p in hull:
q = _next_hull_pt(points, p)
if q != hull[0]:
hull.append(q)
return hull

We will end up going through h iterations of the loop, where h is the number of points on the convex hull. Each iteration takes $$O(n)$$ time to find the furthest right point, so the total time required is $$O(nh)$$, which is suboptimal. However, it is important to remember that suboptimal does not mean it is useless. The Jarvis March can easily handle very large datasets in memory constrained environments. The only information you need to maintain between iterations is the first and last point found. Finding the next point scans through all points only once; order does not matter. These points could come from a database cursor, for instance. In this case, your function would return an iterator of the hull edges (or points) that generates them on demand, each time next() is called.

You can also download the complete version of the code. Next post I’ll cover the Graham Scan.