How to find two most distant points?_问答_开发者

This is a question that 开发者_开发知识库I was asked on a job interview some time ago. And I still can't figure out sensible answer.

Question is:

you are given set of points (x,y). Find 2 most distant points. Distant from each other.

For example, for points: (0,0), (1,1), (-8, 5) - the most distant are: (1,1) and (-8,5) because the distance between them is larger from both (0,0)-(1,1) and (0,0)-(-8,5).

The obvious approach is to calculate all distances between all points, and find maximum. The problem is that it is O(n^2), which makes it prohibitively expensive for large datasets.

There is approach with first tracking points that are on the boundary, and then calculating distances for them, on the premise that there will be less points on boundary than "inside", but it's still expensive, and will fail in worst case scenario.

Tried to search the web, but didn't find any sensible answer - although this might be simply my lack of search skills.

For this specific problem, with just a list of Euclidean points, one way is to find the convex hull of the set of points. The two distant points can then be found by traversing the hull once with the rotating calipers method.

Here is an O(N log N) implementation:

http://mukeshiiitm.wordpress.com/2008/05/27/find-the-farthest-pair-of-points/

If the list of points is already sorted, you can remove the sort to get the optimal O(N) complexity.

For a more general problem of finding most distant points in a graph: Algorithm to find two points furthest away from each other

The accepted answer works in O(N^2).

Boundary point algorithms abound (look for convex hull algorithms). From there, it should take O(N) time to find the most-distant opposite points.

From the author's comment: first find any pair of opposite points on the hull, and then walk around it in semi-lock-step fashion. Depending on the angles between edges, you will have to advance either one walker or the other, but it will always take O(N) to circumnavigate the hull.

You are looking for an algorithm to compute the diameter of a set of points, Diam(S). It can be shown that this is the same as the diameter of the convex hull of S, Diam(S) = Diam(CH(S)). So first compute the convex hull of the set.

Now you have to find all the antipodal points on the convex hull and pick the pair with maximum distance. There are O(n) antipodal points on a convex polygon. So this gives a O(n lg n) algorithm for finding the farthest points.

This technique is known as Rotating Calipers. This is what Marcelo Cantos describes in his answer.

If you write the algorithm carefully, you can do without computing angles. For details, check this URL.

A stochastic algorithm to find the most distant pair would be

Choose a random point
Get the point most distant to it
Repeat a few times
Remove all visited points
Choose another random point and repeat a few times.

You are in O(n) as long as you predetermine "a few times", but are not guaranteed to actually find the most distant pair. But depending on your set of points the result should be pretty good. =)

This question is introduced at Introduction to Algorithm. It mentioned 1) Calculate Convex Hull O(NlgN). 2) If there is M vectex on Convex Hull. Then we need O(M) to find the farthest pair.

I find this helpful links. It includes analysis of algorithm details and program. http://www.seas.gwu.edu/~simhaweb/alg/lectures/module1/module1.html

Wish this will be helpful.

Find the mean of all the points, measure the difference between all points and the mean, take the point the largest distance from the mean and find the point farthest from it. Those points will be the absolute corners of the convex hull and the two most distant points. I recently did this for a project that needed convex hulls confined to randomly directed infinite planes. It worked great.

See the comments: this solution isn't guaranteed to produce the correct answer.

Just a few thoughts:

You might look at only the points that define the convex hull of your set of points to reduce the number,... but it still looks a bit "not optimal".

Otherwise there might be a recursive quad/oct-tree approach to rapidly bound some distances between sets of points and eliminate large parts of your data.

This seems easy if the points are given in Cartesian coordinates. So easy that I'm pretty sure that I'm overlooking something. Feel free to point out what I'm missing!

Find the points with the max and min values of their x, y, and z coordinates (6 points total). These should be the most "remote" of all the boundary points.
Compute all the distances (30 unique distances)
Find the max distance
The two points that correspond to this max distance are the ones you're looking for.

Here's a good solution, which works in O(n log n). It's called Rotating Caliper’s Method. https://www.geeksforgeeks.org/maximum-distance-between-two-points-in-coordinate-plane-using-rotating-calipers-method/

Firstly you find the convex hull, which you can make in O(n log n) with the Graham's scan. Only the point from the convex hull can provide you the maximal distance. This algorithm arranges points of the convex hull in the clockwise traversal. This property will be used later.

Secondly, for all the points on the convex hull, you'll need to find the most distant point on this hull (it's called the antipodal point here). You don't have to find all the antipodal points separately (which would give quadratic time). Let's say the points of the convex hall are called p_1, ..., p_n, and their order corresponds to the clockwise traversal. There is a property of convex polygons that when you iterate through points p_j on the hull in the clockwise order and calculate the distances d(p_i, p_j), these distances firstly don't decrease (and maybe increase) and then don't increase (and maybe decrease). So you can find the maximum distance easily in this case. But when you've found the correct antipodal point p_j* for the p_i, you can start this search for p_{i+1} with the candidates points starting from that p_j*. You don't need to check all previously seen points. in total p_i iterates through points p_1, ..., p_n once, and p_j iterates through these points at most twice, because p_j can never catch up p_i as it would give zero distance, and we stop when the distance starts decreasing.

A solution that has runtime complexity O(N) is a combination of the above answers. In detail: (1) One can compute the convex hull with runtime complexity O(N) if you use counting sort as an internal polar angle sort and are willing to use angles rounded to the nearest integer [0, 359], inclusive.

(2) Note that the number of points on the convex hull is then N_H which is usually less than N.
We can speculate about the size of the hull from information in Cormen et al. Introduction to Algorithms, Exercise 33-5. For sparse-hulled distributions of a unit-radius disk, a convex polygon with k sides, and a 2-D normal distribution respectively as n^(1/3), log_2(n), sqrt(log_2(n)).

The furthest pair problem is then between comparison of points on the hull. This is N_H^2, but each leading point's search for distance point can be truncated when the distances start to decrease if the points are traversed in the order of the convex hull (those points are ordered CCW from first point). The runtime complexity for this part is then O(N_H^2).

Because N_H^2 is usually less than N, the total runtime complexity for furthest pair is O(N) with a caveat of using integer degree angles to reduce the sort in the convex hull to linear.

Given a set of points {(x1,y1), (x2,y2) ... (xn,yn)} find 2 most distant points.

My approach:

1). You need a reference point (xa,ya), and it will be:
xa = ( x1 + x2 +...+ xn )/n
ya = ( y1 + y2 +...+ yn )/n

2). Calculate all distance from point (xa,ya) to (x1,y1), (x2,y2),...(xn,yn)
The first "most distant point" (xb,yb) is the one with the maximum distance.

3). Calculate all distance from point (xb,yb) to (x1,y1), (x2,y2),...(xn,yn)
The other "most distant point" (xc,yc) is the one with the maximum distance.

So you got your most distant points (xb,yb) (xc,yc) in O(n)

For example, for points: (0,0), (1,1), (-8, 5)

1). Reference point (xa,ya) = (-2.333, 2)

2). Calculate distances:
from (-2.333, 2) to (0,0) : 3.073
from (-2.333, 2) to (1,1) : 3.480
from (-2.333, 2) to (-8, 5) : 6.411
So the first most distant point is (-8, 5)

3). Calculate distances:
from (-8, 5) to (0,0) : 9.434
from (-8, 5) to (1,1) : 9.849
from (-8, 5) to (-8, 5) : 0
So the other most distant point is (1, 1)