Below given a point set in the rectilinear metric (the height/width of any cell=
ID: 3812498 • Letter: B
Question
Below given a point set in the rectilinear metric (the height/width of any cell=1) where the closest pair of points should be found using divide and conquer. Show - the first partition of the point set (draw a line) -the closest pair in the left part (connect solid), left = _______ , and the right part (connect solid), right = ______ - the middle strip (shade) - pairs in the middle strip for which distances should be computed (connect dashed) - closest pair in the middle strip (connect solid) 4. Below given a point set in the rectilinear metric the height/width of any cells 1) pair of points should be found using divide and conquer. Show where the closest the first partition of the point set (draw a line) -the closest pair in the left part (connect solid), olen and the right part (connect solid), right the middle strip (shade) pairs in the middle strip for which distances should be computed (connect dashed) closest pair in the middle strip (connect solid)Explanation / Answer
1D problem can be solved in O(n log n) via sorting. • Sorting, however, does not generalize to higher dimensions. So, let’s develop a divide-and-conquer for 1D. • Divide the points S into two sets S1, S2 by some x-coordinate so that p < q for all p S1 and q S2. • Recursively compute closest pair (p1, p2) in S1 and (q1, q2) in S2.
The closest pair is {p1, p2}, or {q1, q2}, or some {p3, q3} where p3 S1 and q3 S2. • Key Observation: If m is the dividing coordinate, then p3, q3 must be within of m. • In 1D, p3 must be the rightmost point of S1 and q3 the leftmost point of S2, but these notions do not generalize to higher dimensions. • How many points of S1 can lie in the interval (m , m]? • By definition of , at most one. Same holds for S2.
We partition S into S1, S2 by vertical line ` defined by median x-coordinate in S. • Recursively compute closest pair distances 1 and 2. Set = min(1, 2). • Now compute the closest pair with one point each in S1 and S2.
Given n points with -sparsity condition, find all pairs within distance . • Divide the set into S1, S2 by a median place H. Recursively solve the problem in two halves. • Project all points lying within thick slab around H onto H. Call this set S 0 . • S 0 inherits the -sparsity condition. Why?. • Recursively solve the problem for S 0 in d 1 space. • The algorithms satisfies the recurrence U(n, d) = 2U(n/2, d) + U(n, d 1) + O(n). which solves to U(n, d) = O(n(log n) d1 ).
#include <stdio.h>
#include <float.h>
#include <stdlib.h>
#include <math.h>
// A structure to represent a Point in 2D plane
struct Point
{
int x, y;
};
/* Following two functions are needed for library function qsort().
Refer: http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
// Needed to sort array of points according to X coordinate
int compareX(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->x - p2->x);
}
// Needed to sort array of points according to Y coordinate
int compareY(const void* a, const void* b)
{
Point *p1 = (Point *)a, *p2 = (Point *)b;
return (p1->y - p2->y);
}
// A utility function to find the distance between two points
float dist(Point p1, Point p2)
{
return 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
float bruteForce(Point P[], int n)
{
float min = FLT_MAX;
for (int i = 0; i < n; ++i)
for (int j = i+1; j < n; ++j)
if (dist(P[i], P[j]) < min)
min = dist(P[i], P[j]);
return min;
}
// A utility function to find minimum of two float values
float min(float x, float y)
{
return (x < y)? x : y;
}
// 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
float stripClosest(Point strip[], int size, float d)
{
float min = d; // Initialize the minimum distance as d
qsort(strip, size, sizeof(Point), compareY);
// 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 < size; ++i)
for (int j = i+1; j < size && (strip[j].y - strip[i].y) < min; ++j)
if (dist(strip[i],strip[j]) < min)
min = dist(strip[i], strip[j]);
return min;
}
// A recursive function to find the smallest distance. The array P contains
// all points sorted according to x coordinate
float closestUtil(Point P[], int n)
{
// If there are 2 or 3 points, then use brute force
if (n <= 3)
return bruteForce(P, n);
// Find the middle point
int mid = n/2;
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
float dl = closestUtil(P, mid);
float dr = closestUtil(P + mid, n-mid);
// Find the smaller of two distances
float d = min(dl, dr);
// Build an array strip[] that contains points close (closer than d)
// to the line passing through the middle point
Point strip[n];
int j = 0;
for (int i = 0; i < n; i++)
if (abs(P[i].x - midPoint.x) < d)
strip[j] = P[i], j++;
// Find the closest points in strip. Return the minimum of d and closest
// distance is strip[]
return min(d, stripClosest(strip, j, d) );
}
// The main functin that finds the smallest distance
// This method mainly uses closestUtil()
float closest(Point P[], int n)
{
qsort(P, n, sizeof(Point), compareX);
// Use recursive function closestUtil() to find the smallest distance
return closestUtil(P, n);
}
// Driver program to test above functions
int main()
{
Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}};
int n = sizeof(P) / sizeof(P[0]);
printf("The smallest distance is %f ", closest(P, n));
return 0;
}
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