Question

I am trying to make a closest pair alg using div and conq.

using this base for the alg please show me how I would go about doing this. In C++ obviously.

std::pair<Point<T>, Point<T> > closeP( vector<Point<T> > v){

return std::pair<Point<T>, Point<T> > (v[0],v[1]);

}

Here is a brute force alg that can be used to create the base cases for the div and conq alg.
std::pair<Point<T> ,Point<T> > closePBF( vector<Point<T> > pts){
T min = dist(pts[0],pts[1]);
size_t min1 = 0;
size_t min2 = 1;

for(size_t i = 0; i< pts.size()-1; ++i){
for(size_t j = i+1; j <pts.size(); ++j){
T d = dist(pts[i],pts[j]);
if (d<min) {
min = d;
min1 = i;
min2 = j;
}
}
}

I have a tester so I dont need anything extra to see that it works. Both algs (the brute force and the base of the one I want created) are using a template with typename<T>. Try to keep the alg fairly simple.

Answer 1

C++ code for the above problem is as follows-

 #include <bits/stdc++.h> using namespace std; // A structure to represent a Point in 2D plane class Point { public: int x, y; }; // 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 function 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 code int main() { Point P[] = {{2, 3}, {12, 30}, {40, 50}, {5, 1}, {12, 10}, {3, 4}}; int n = sizeof(P) / sizeof(P[0]); cout << "The smallest distance is " << closest(P, n); return 0; } 

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