分治法 Divide and Conquer - Closest Pair of Points 找最近点对
分治法找最近点对，可以把时间复杂度从暴力的O(n^2)降低到O(nlogn)
原理就是把图中的所有点分为左右两半，接着递归再二分左右两半。直到区间内只有两个点，易得两点间长度。
之后是合并的过程，从左右区间中取出更小的那一个。接着，比较中间区域strip内的点对，看看有没有更短的点对。（这个过程经过证明最多只要比较7次！）
最后得到最小的那个长度。
We are given an array of n points in the plane, and the problem is to find out the closest pair of points in the array. This problem arises in a number of applications.For example, in air-traffic control, you may want to monitor
planes that come too close together, since this may indicate a possible collision. Recall the following formula for distance between two points p and q.
The Brute force solution is O(n^2), compute the distance between each pair and return the smallest. We can calculate the smallest distance in O(nLogn) time using Divide and Conquer strategy. In this post, a O(n x (Logn)^2) approach is discussed. We will be discussing a O(nLogn) approach in a separate post.
Algorithm
Following are the detailed steps of a O(n (Logn)^2) algortihm.
Input:An array of n pointsP[]
Output:The smallest distance between two points in the given array.
As a pre-processing step, input array is sorted according to x coordinates.
1)Find the middle point in the sorted array, we can takeP[n/2]as middle point.
2)Divide the given array in two halves. The first subarray contains points from P[0] to P[n/2]. The second subarray contains points from P[n/2+1] to P[n-1].
3)Recursively find the smallest distances in both subarrays. Let the distances be dl and dr. Find the minimum of dl and dr. Let the minimum be d.
4)From above 3 steps, we have an upper bound d of minimum distance. Now we need to consider the pairs such that one point in pair is from left half and other is from right half. Consider the vertical line passing through passing through P[n/2] and find all points whose x coordinate is closer than d to the middle vertical line. Build an array strip[] of all such points.
5)Sort the array strip[] according to y coordinates. This step is O(nLogn). It can be optimized to O(n) by recursively sorting and merging.
6)Find the smallest distance in strip[]. This is tricky. From first look, it seems to be a O(n^2) step, but it is actually O(n). It can be proved geometrically that for every point in strip, we only need to check at most 7 points after it (note that strip is sorted according to Y coordinate). Seethisfor more analysis.
7)Finally return the minimum of d and distance calculated in above step (step 6)
Time ComplexityLet Time complexity of above algorithm be T(n). Let us assume that we use a O(nLogn) sorting algorithm. The above algorithm divides all points in two sets and recursively calls for two sets. After dividing, it finds the strip
in O(n) time, sorts the strip in O(nLogn) time and finally finds the closest points in strip in O(n) time. So T(n) can expressed as follows
T(n) = 2T(n/2) + O(n) + O(nLogn) + O(n)
T(n) = 2T(n/2) + O(nLogn)
T(n) = T(n x Logn x Logn)
Notes
1)Time complexity can be improved to O(nLogn) by optimizing step 5 of the above algorithm. We will soon be discussing the optimized solution in a separate post.
2)The code finds smallest distance. It can be easily modified to find the points with smallest distance.
3)The code uses quick sort which can be O(n^2) in worst case. To have the upper bound as O(n (Logn)^2), a O(nLogn) sorting algorithm like merge sort or heap sort can be used
用Java重写了一遍：
package CloestPairOfPoints;
import java.util.Arrays;
import java.util.Comparator;
/**
* Divide and Conquer to get the cloest pair of points
*
* O(nlogn) time complexity
*
*/
public class CloestPairOfPoints {
public static void main(String[] args) {
Point[] points = new Point[6];
points[0] = new Point(2, 3);
points[1] = new Point(12, 30);
points[2] = new Point(40, 50);
points[3] = new Point(5, 1);
points[4] = new Point(12, 10);
points[5] = new Point(3, 4);
double minDist = closest(points);
System.out.println(minDist);
double minDistBruteForce = bruteForce(points);
System.out.println(minDistBruteForce);
}
// A utility function to find the distance between two points
private static double dist(Point p1, Point p2) {
return Math.sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y)
* (p1.y - p2.y));
}
// A Brute Force method to return the smallest distance between two points
// in P[] of size n
public static double bruteForce(Point[] P) {
double min = Double.MAX_VALUE;
int n = P.length;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
min = Math.min(min, dist(P[i], P[j]));
}
}
return min;
}
// The main functin that finds the smallest distance
// This method mainly uses closestUtil()
public static double closest(Point[] P) {
System.out.println("Before soring:");
System.out.println(Arrays.toString(P));
Arrays.sort(P, new Comparator<Point>() {
@Override
public int compare(Point p1, Point p2) {
return p1.x - p2.x;
}
});
System.out.println("After sorting in x axis:");
System.out.println(Arrays.toString(P));
return closestUtil(P, 0, P.length - 1);
}
// A recursive function to find the smallest distance. The array P contains
// all points sorted according to x coordinate
private static double closestUtil(Point[] P, int begin, int end) {
int n = end - begin + 1;
if (n <= 3) { // If there are 2 or 3 points, then use brute force
return bruteForce(P);
}
int mid = n / 2; // Find the middle point
Point midPoint = P[mid];
// Consider the vertical line passing through the middle point
// calculate the smallest distance dl on left of middle point and
// dr on right side
double left = closestUtil(P, begin, mid);
double right = closestUtil(P, mid, end);
// Find the smaller of two distances
double distMin = Math.min(left, right);
Point[] strip = new Point[n];
int stripLen = 0;
// Build an array strip[] that contains points close (closer than d)
// to the line passing through the middle point
for (int i = 0; i < n; i++) {
if (Math.abs(P[i].x - midPoint.x) < distMin) {
strip[stripLen] = P[i];
stripLen++;
}
}
// Find the closest points in strip. Return the minimum of d and closest
// distance is strip[]
return Math.min(distMin, stripCloest(strip, stripLen, distMin));
}
// A utility function to find the distance beween the closest points of
// strip of given size. All points in strip[] are sorted accordint to
// y coordinate. They all have an upper bound on minimum distance as d.
// Note that this method seems to be a O(n^2) method, but it's a O(n)
// method as the inner loop runs at most 6 times
private static double stripCloest(Point[] strip, int stripLen,
double distMin) {
Arrays.sort(strip, 0, stripLen, new Comparator<Point>() {
@Override
public int compare(Point p1, Point p2) {
return p1.y - p2.y;
}
});
System.out.println("After sorting in y axis for strip:");
System.out.println(Arrays.toString(strip));
double min = distMin;
// O(C*n) time complexity, not O(n^2)
// Pick all points one by one and try the next points till the difference
// between y coordinates is smaller than d.
// This is a proven fact that this loop runs at most 6 times
for (int i=0; i<stripLen; i++) {
for(int j=i+1; j<stripLen && (strip[j].y - strip[i].y < min); j++) {
min = Math.min(min, dist(strip[i], strip[j]));
}
}
return min;
}
static class Point {
int x, y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return "[x=" + x + ", y=" + y + "]";
}
}
}
Ref：
http://users.soe.ucsc.edu/~skourtis/201/open/lecture3-ClosestPair.pdf
http://www.geeksforgeeks.org/closest-pair-of-points/