Gift wrapping algorithm

Two dimensions

	Sorry, but you need Java to see the animation.	Find least point A (with minimum y coordinate) as a starting point.
	Sorry, but you need Java to see the animation.	We can find B where all points lie to the left of AB by scanning through all the points.
	Sorry, but you need Java to see the animation.	Similarly, we can find C where all points lie to the left of BC. We can repeat this to find the next point and so on.

If the number of sides of the hull is h then the complexity is nh.

Three dimensions

	Sorry, but you need Java to see the animation.	Find a starting edge AB by using the 2D algorithm on the projection of the points on the XY plane.
	Sorry, but you need Java to see the animation.	We can find C where all points lie to the left of triangle ABC by scanning through all the points. The view to the left is looking along AB so that the triangle appears as a line.
	Sorry, but you need Java to see the animation.	Similarly, we can find D where all points lie to the left of ACD. The view to the left is looking along AC so that the triangle appears as a line. We also need to find the other triangle adjacent to BA and CB, So we need a stack or a queue to organise the search. Can repeat this to find the next point and so on.

More dimensions