Thread: finding nearest neighbour distance

and the number of points that I am dealing with will be of the order of N= 1000 say, so not very large,,

the first bit of my code is as follows so far:

insert

Code:

struct point{
	double x;
	double y;
}; 
typedef struct point point;

double distance (point a, point b); 

double distancex (point a, point b);

double distancey (point a, point b);

int main(){
int i = 0, j = 0;
double entropyxy, entropyx, entropyy;
double dis, result;
double disx, resultx;
double disy, resulty;
double digammaone, digammaN;
double mutual_information;


point myPoints[3] = {{1.0, 1.0}, {2.0, 3.0}, {3.0, 4.0}};
	 
	 
	
	 
		for (i = 0; i < 3; i++){
		

		
		result = distance(myPoints[i], myPoints[i+1]);
		printf("first distance:%f\n", result);
		
		for (j = 0; j < 3; j++){
		
		
		
			if (i == j)
			 continue;
				
				
				
           dis = distance(myPoints[i], myPoints[j]);
            printf("%d %d\n", i, j);
            printf("%f\n", dis);
				
           if (result > dis)
           result = dis;
		 
			
          
	}
	
		   printf("shortest: %f\n", result);
	    
		   if(result != 0){entropyxy += (log(2*result));}
		   printf("entropyxy: %f\n", entropyxy);
		   
	
	}

what I need help with is the first bit. As it stands I have only worked with 3 points. But what I want to do is use a data file as my input. This data file contains two columns: first is x values, the second column are all the y values. Each row thus makes 1 point.

So I have a N x 2 array in my input file, where N is the number of points. How do I use this as my input rather than just the 3 points I have in my code at the moment? I understand that I will have to loop over N, and scanf each in but am having trouble with the syntax.

Any help would be appreciated,

Thanks

zesty

Originally Posted by brewbuck

Many points are so far away that you do not have to calculate the actual distance to prove that the point is not the closest point.

Suppose you have the current closest distance to a neighboring point in variable d. Then, if you are considering a new point with x distance dx and y distance dy, then if dx >= d or dy >= d, you know that this point cannot be closer, and you can avoid squaring numbers and taking square roots.

There is a reason behind squaring. Will your solution hold true for all points across any quadrant?