On Thursday, July 11, 2013 10:50:27 PM UTC-7, quasi wrote: > I'll try to explain in more detail ...
Let me say first that I really appreciate the time you've taken to help me with this problem. But I am a little dense when it comes to proofs, so please bear with me.
> (8) Since L0 does not pass through any of the remaining N-1 > points, a sufficiently small clockwise rotation of the line L0 > through the point (x_k,y_k) will still yield a lower bounding > line. Similarly, a sufficiently small counterclockwise rotation > of the line L0 through the point (x_k,y_k) will also still yield > a lower bounding line.
This I don't understand. Imagine three points forming an isosceles triangle, A---B \ / C and a line L0 passing through C, but not either A or B (for example, parallel to AB). Rotation of this line doesn't change the sum of the distances from L0 to A and L0 to B, therefore rotation doesn't produce a lower bounding line.
If the triangle in this example is not isosceles, rotation in one direction will increase the sum and the other will decrease it.
My data has some additional constraints, but your proof seemed to be for the general case.