Reputation: 759
I have 2 lists containing points (x,y and z) and would like to find the closest points.
I assume I need to do something like this:
for (int i = 0; i < myarray1.Count; i++)
{
for (int j = 0; j < myarray2.Count; j++)
{
// Calculate the quadratic distance between 2 points (point in index i and j)
// Then store the minimum distance I guess?
}
}
Upvotes: 2
Views: 5006
Reputation: 493
Formula is
Math.sqrt(Math.pow(Math.abs(x1-x2),2) + Math.pow(Math.abs(y1-y2),2)+ Math.pow(Math.abs(z1-z2),2))
Upvotes: 0
Reputation: 10172
To compute the distance:
double sqr(double x) {return x*x;}
double distance(MyPoint a, MyPoint b) {
return sqrt(sqr(a.x-b.x)+sqr(a.y-b.y)+sqr(a.z-b.z);
}
Then in the second loop you store the minimum distance found so far:
double d = distance(myarray1[i],myarray2[j]);
if (d<min_d) min_d = d;
where min_d is defined at the beginning:
double min_d = Float.MAX_VALUE;
Upvotes: 1
Reputation: 36340
In C# I would do this using Linq.
First I would define the function that calculates the distance between two points as in Emanuelle Paolini's answer:
public double Distance(Point p1, Point p2)
{
return Math.Sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p1.y) * (p1.y - p1.y) + (p1.z - p1.z) * (p1.z - p1.z));
}
Then I would query the two lists as follows:
var distanceQuery = from p1 in myarray1
from p2 in myarray2
select Dist(p1, p2);
And finally I would retrieve the minimum distance:
var minimumDistance = distanceQuery.Min();
Upvotes: 0
Reputation: 4483
another option is to use Kd-tree
using the Nearest neighbour search will give you a O(log n)
complexity to find the nearest point to a given set of points, and your code will have O( n log n)
, instead of O (n^2)
.
see here for implementation and example of how to use it.
Upvotes: 4
Reputation: 4118
double min_dist = DOUBLE_MAX;
for (i = 0; i < myarray1.Count; i++)
{
for (j = 0; j < myarray2.Count; j++)
{
curr_dist = dist(a[i],a[j]);
if( min_dist > curr_dist)
{
min_dist = curr_dist;
}
}
}
where
double dist(Point a, Point b) {
return sqrt(pow(a.x-b.x,2)+pow(a.y-b.y,2)+pow(a.z-b.z,2);
}
Upvotes: 2