- Start at some extreme point, which is guaranteed to be on the hull.
- At each step, test
*each*of the points, and find the one which makes the largest right-hand turn. That point has to be the next one on the hull.

As we can see, the algorithm is not very fast in this case.
Quick-hull would probably be faster. In fact
if *n* points are arranged in a circle, Jarvis' march will
take time proportional to *n^2*, as you can see in the following
demo:

If you think about it, Jarvis' march takes time proportional to
*nh*, where *n* is the number of input points, and
*h* is the number of points on the hull. In other words,
Jarvis' march is *output-sensitive*. The algorithm works best
for inputs such as the following, where *h* is 3:

Alejo Hausner, CS Department, Princeton University Last modified: Wed Jul 17 14:47:54 1996