Hi Guys,

I was wondering if someone could help me with this small program, i am kinda new to programming and well i have been given this assignment on collision detection for airplanes. i have done some coding implementing the algorithm and well i have specified the coordinates or points that would represent the planes and thus the program recursively calculates the shortest distance between these points, however i need the points to be moving at a specific path such that the program can detect a collision, and also i need to have random points everytime the program runs..................

I really need help.....

Here is the code that i have done for now......

```
{
// A divide and conquer program in C++ to find the smallest distance from a
// given set of points.
#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;
};
// 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[] = {{4, 3}, {20, 30}, {10, 50}, {5, 10}, {1, 10}, {30, 4}};
int n = sizeof(P) / sizeof(P[0]);
printf("The smallest distance is %f ", closest(P, n));
system("PAUSE");
return 0;
}
}
```