How many integer points within the three points forming a triangle?

Question

Actually this is a classic problem as SO user Victor put it (in another SO question regarding which tasks to ask during an interview).

I couldn\'t do it in an hour

Answer 1

I have this idea -

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle. Let 'count' be the number of integer points forming the triangle.

If we need the points on the triangle edges then using Euclidean Distance formula http://en.wikipedia.org/wiki/Euclidean_distance, the length of all three sides can be ascertained. The sum of length of all three sides - 3, would give that count.

To find the number of points inside the triangle we need to use a triangle fill algorithm and instead of doing the actual rendering i.e. executing drawpixel(x,y), just go through the loops and keep updating the count as we loop though. A triangle fill algorithm from

Fundamentals of Computer Graphics by Peter Shirley,Michael Ashikhmin

should help. Its referred here http://www.gidforums.com/t-20838.html

cheers

Answer 2

I found a quite useful link which clearly explains the solution to this problem. I am weak in coordinate geometry so I used this solution and coded it in Java which works (at least for the test cases I tried..)

http://mathforum.org/library/drmath/view/55169.html

public int points(int[][] vertices){
      int interiorPoints = 0;
      double triangleArea = 0;
      int x1 = vertices[0][0], x2 = vertices[1][0], x3 = vertices[2][0];
      int y1 = vertices[0][1], y2 = vertices[1][1], y3 = vertices[2][1];

      triangleArea = Math.abs(((x1-x2)*(y1+y2))                             
                + ((x2-x3)*(y2+y3))
                + ((x3-x1)*(y3+y1)));

      triangleArea /=2;
      triangleArea++;

      interiorPoints = Math.abs(gcd(x1-x2,y1-y2))
                + Math.abs(gcd(x2-x3, y2-y3))
                + Math.abs(gcd(x3-x1, y3-y1));

      interiorPoints /=2;

      return  (int)(triangleArea - interiorPoints);
 }

Answer 3

I only have half an answer for a non-brute-force method. If the vertices were integer, you could reduce it to figuring out how to find how many integer points the edges intersect. With that number and the area of the triangle (Heron's formula), you can use Pick's theorem to find the number of interior integer points.

Edit: for the other half, finding the integer points that intersect the edge, I suspect that it's the greatest common denominator between the x and y difference between the points minus one, or if the distance minus one if one of the x or y differences is zero.

Answer 4

Here is a Python implementation of @Prabhala's solution:

from collections import namedtuple
from fractions import gcd


def get_points(vertices):
    Point = namedtuple('Point', 'x,y')
    vertices = [Point(x, y) for x, y in vertices]

    a, b, c = vertices

    triangle_area = abs((a.x - b.x) * (a.y + b.y) + (b.x - c.x) * (b.y + c.y) + (c.x - a.x) * (c.y + a.y))
    triangle_area /= 2
    triangle_area += 1

    interior = abs(gcd(a.x - b.x, a.y - b.y)) + abs(gcd(b.x - c.x, b.y - c.y)) + abs(gcd(c.x - a.x, c.y - a.y))
    interior /= 2

    return triangle_area - interior

Usage:

print(get_points([(-1, -1), (1, 0), (0, 1)]))  # 1
print(get_points([[2, 3], [6, 9], [10, 160]]))  # 289

Answer 5

Assuming the vertices are at integer coordinates, you can get the answer by constructing a rectangle around the triangle as explained in Kyle Schultz's An Investigation of Pick's Theorem.

For a j x k rectangle, the number of interior points is

I = (j – 1)(k – 1).

For the 5 x 3 rectangle below, there are 8 interior points.

_{(source: uga.edu)}

For triangles with a vertical leg (j) and a horizontal leg (k) the number of interior points is given by

I = ((j – 1)(k – 1) - h) / 2

where h is the number of points interior to the rectangle that are coincident to the hypotenuse of the triangles (not the length).

_{(source: uga.edu)}

For triangles with a vertical side or a horizontal side, the number of interior points (I) is given by

_{(source: uga.edu)}

where j, k, h1, h2, and b are marked in the following diagram

_{(source: uga.edu)}

Finally, the case of triangles with no vertical or horizontal sides can be split into two sub-cases, one where the area surrounding the triangle forms three triangles, and one where the surrounding area forms three triangles and a rectangle (see the diagrams below).

The number of interior points (I) in the first sub-case is given by

_{(source: uga.edu)}

where all the variables are marked in the following diagram

_{(source: uga.edu)}

The number of interior points (I) in the second sub-case is given by

_{(source: uga.edu)}

where all the variables are marked in the following diagram

_{(source: uga.edu)}

Answer 6

Quick n'dirty pseudocode:

-- Declare triangle
p1 2DPoint = (x1, y1);
p2 2DPoint = (x2, y2);
p3 2DPoint = (x3, y3);
triangle [2DPoint] := [p1, p2, p3];

-- Bounding box
xmin float = min(triangle[][0]);
xmax float = max(triangle[][0]);
ymin float = min(triangle[][1]);
ymax float = max(triangle[][1]);

result [[float]];

-- Points in bounding box might be inside the triangle
for x in xmin .. xmax {
  for y in ymin .. ymax {
    if a line starting in (x, y) and going in any direction crosses one, and only one, of the lines between the points in the triangle, or hits exactly one of the corners of the triangle {
      result[result.count] = (x, y);
    }
  }
}