找到与平面中所有其他点不共线的点

Question

Given a list of N points in the plane in general position (no three are collinear), find a new point p that is not collinear with any pair of the N original points.

We obviously cannot search for every point in the plane, I started with finding the coincidence point of all the lines that can be formed with the given points, or making a circle with them something.. I dont have any clue how to check all the points.

Question found in http://introcs.cs.princeton.edu/java/42sort/

I found this question in a renowned algorithm book that means it is answerable, but I cannot think of an optimal solution, thats why I am posting it here so that if some one knows it he/she can answer it

Answer 1

The best I can come up with is an N^2 algorithm. Here goes:

1. Choose a tolerance e to control how close you're willing to come to a line formed from the points in the set.
2. Compute the convex hull of your set of points.
3. Choose a line L parallel to one of the sides of the convex hull, at a distance 3e outside the hull.
4. Choose a point P on L, so that P is outside the projection of the convex hull on L. The projection of the convex hull on L is an interval of L. P must be placed outside this interval.
5. Test each pair of points in the set. For a particular line M formed by the 2 test points intersects a disc of radius 2e around P, move P out further along L until M no longer intersects the disc. By the construction of L, there can be no line intersecting the disk parallel to L, so this can always be done.
6. If M crosses L beyond P, move P beyond that intersection, again far enough that M doesn't pass through the disc.
7. After all this is done, choose your point at distance e , on the perpendicular to L at P. It can be colinear with no line of the set.

I'll leave the details of how to choose the next position of P along L in step 5 to you,

There are some obvious trivial rejection tests you can do so that you do more expensive checks only with the test line M is "parallel enough" to L.

Finally, I should mention that it is probably possible to push P far enough out that numerical problems occur. In that case the best I can suggest is to try another line outside of the convex hull by a distance of at least 3e .

Answer 2

You can actually solved it using a simple O(nlogn) algorithm, which we will then improve to O(n). Name A the bottom most point (in case of tie choose the one that is has smaller x coordinate). You can now sort in clockwise order the rest of the points using the CCW. Now as you process each point from the sorted order you can see that between any two successive points having different angle with point A and the bottom axis (let these be U, V) there is no point having angle c, with U <= c <= V. So we can add any point in this section and it is guaranteed that it won't be collinear with any other points from the set. So, all you need is to find one pair of adjacent points and you are done. So, find the minimum and the second minimum angle with A (these should be different) in O(n) time and select any point in between them.