CODEWITHSUNDEEP
c#
Headpuster
Headpuster

Reputation: 267

Algorithm for calculating a distance between 2 3-dimensional points?

I have two 3-D points.
Example:

float[] point1 = new float[3] {1.3919023, 6.12837912, 10.391283};
float[] point2 = new float[3] {48.3818, 38.38182, 318.381823};

Anyone an idea for an algorithm to calculate the distance in float between the points?

Upvotes: 8

Views: 21409

Answers (5)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 727067

In C# with LINQ you can do this:

var dist = Math.Sqrt(point1.Zip(point2, (a, b) => (a - b)*(a - b)).Sum());

This sums up squared pairwise differences between individual coordinates, and returns the arithmetic square root of the sum.

EDIT : This solution works for any number of dimensions greater or equal to one (thanks to Austin Salonen for pointing it out).

Upvotes: 21

Ron Warholic
Ron Warholic

Reputation: 10074

The Euclidean distance between two 3D points is:

float deltaX = x1 - x0;
float deltaY = y1 - y0;
float deltaZ = z1 - z0;

float distance = (float) Math.Sqrt(deltaX * deltaX + deltaY * deltaY + deltaZ * deltaZ);

And in N dimensions (untested and vulnerable to overflow):

float DistanceN(float[] first, float[] second) {
  var sum = first.Select((x, i) => (x - second[i]) * (x - second[i])).Sum();
  return Math.Sqrt(sum);
}

Edit: I much prefer the Zip solution posted by dasblinkenlight below!

Upvotes: 26

Farid Movsumov
Farid Movsumov

Reputation: 12735

enter image description here

float distance=(float) Math.Sqrt(Math.Pow(point1[0]-point2[0],2) + Math.Pow(point1[1]-point2[1],2) + Math.Pow(point1[2]-point2[2],2))

Upvotes: 10

Marco
Marco

Reputation: 57593

If you have two points:
P1 = (x1, y1, z1)
P2 = (x2, y2, z2)
distance is SQRT((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)

So you could use

float deltax = point2[0] - point1[0];
float deltay = point2[1] - point1[1];
float deltaz = point2[2] - point1[2];
float distance = (float) Math.Sqrt(
    (deltax * deltax) +
    (deltay * deltay) +
    (deltaz * deltaz));

Upvotes: 2

InBetween
InBetween

Reputation: 32790

Like 2D but with one more coordinate:

P1(x1, y1, z1); P2(x2, y2, z2)

d = SquareRootOf((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2) + (z1-z2)*(z1-z2));

Obviously not written down in C# but you get the idea.

Upvotes: 4

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