Triangle bounding box

Question

I'm supposed to write code that calculates the bounding box of a triangle. The bounding box coordinates should be written to

triangle->bx, triangle->by, triangle->bw, triangle->bh

where

bx, by is the upper left corner of the box
bw, bh is the width and height of the box

Do I treat my points as coordinates or should I choose a more geometry-based solution?

I tried finding the minimum and maximum values for every coordinate, but it didn't work. Any help would be much appreciated!

if (triangle->sx1 <= triangle->sx2 <= triangle->sx3)
{
    triangle->bx = triangle->sx1;
}
else if (triangle->sx2 <= triangle->sx1 <= triangle->sx3)
{
    triangle->bx = triangle->sx2;
}
else (triangle->bx = triangle->sx3);

if (triangle->sy1 <= triangle->sy2 <= triangle->sy3)
{
    triangle->by = triangle->sy1;
}
else if (triangle->sy2 <= triangle->sy1 <= triangle->sy3)
{
    triangle->by = triangle->sy2;
}
else (triangle->by = triangle->sy3);

if (triangle->sx1 >= triangle->sx2 >= triangle->sx3)
{
    triangle->bw = triangle->sx1;
}
else if (triangle->sx2 >= triangle->sx1 >= triangle->sx3)
{
    triangle->bw = triangle->sx2;
}
else (triangle->bw = triangle->sx3);

if (triangle->sy1 >= triangle->sy2 >= triangle->sy3)
{
    triangle->bh = triangle->sy1;
}
else if (triangle->sy2 >= triangle->sy1 >= triangle->sy3)
{
    triangle->bh = triangle->sy2;
}
else (triangle->bh = triangle->sy3);

You cannot make a triple comparison that way, you must compare them in pairs. — Weather Vane, Oct 11 '16 at 09:28
OK. Guess there has to be a better way of doing that than if-testing for every single case. — user6995957, Oct 11 '16 at 09:30
You have to if-test for every case, but using the proper language syntax. — Weather Vane, Oct 11 '16 at 09:34
Otherwise your idea of finding minimum and maximum values is the way to go, just make the comparisons correctly. — Sami Kuhmonen, Oct 11 '16 at 09:37
A triangle has much more than one bounding box once you allow the box to rotate. — tofro, Oct 11 '16 at 10:24
min3 is min2(min2(a.b),c). The question is whether it is worth writing a little function or macro or just putting the logic inline. — Malcolm McLean, Oct 11 '16 at 10:51

score 4 · Answer 1 · answered Oct 11 '16 at 09:37

To find the bounds of the box containing a triangle, you simply need to find the smallest and largest x and y coordinates from the three coordinates making up the triangle. You can do the comparisons using ternary expressions, which makes the code a bit less ugly. In the code below, I port the x and y triangle coordinates into separate variables so the ternary expression can be more easily read.

int sx1 = triangle->sx1;
int sx2 = triangle->sx2;
int sx3 = triangle->sx3;
int sy1 = triangle->sy1;
int sy2 = triangle->sy2;
int sy3 = triangle->sy3;

int xmax = sx1 > sx2 ? (sx1 > sx3 ? sx1 : sx3) : (sx2 > sx3 ? sx2 : sx3);
int ymax = sy1 > sy2 ? (sy1 > sy3 ? sy1 : sy3) : (sy2 > sy3 ? sy2 : sy3);
int xmin = sx1 < sx2 ? (sx1 < sx3 ? sx1 : sx3) : (sx2 < sx3 ? sx2 : sx3);
int ymin = sy1 < sy2 ? (sy1 < sy3 ? sy1 : sy3) : (sy2 < sy3 ? sy2 : sy3);

triangle->bx = xmin;
triangle->by = ymax;
triangle->bw = xmax - xmin;
triangle->bh = ymax - ymin;

score 0 · Answer 2 · answered Jun 02 '21 at 21:00

0

i might be late on the question but if you have vectorization you can do :

vec3 v1;
vec3 v2;
vec3 v3;

aabb->min = min(v1,min(v2,v3));
aabb->max = max(v1,max(v2,v3));

answered Jun 02 '21 at 21:00

Cewein

396
2
14

Triangle bounding box

2 Answers2