Numerical stability of point-in-triangle test with barycentric coordinates

问题

While looking at various methods for point-in-triangle testing (2D case), I found that the method which uses barycentric coordinates is the most used one. Here is a StackOverflow answer which explains it.

Why is this method the most preferred one? It probably has to do with doing less calculations, but what about numerical stability? Is this algorithm better suited than say, the "same side" technique, for cases in which the point is particularly near the border?

回答1:

If you solve it:

p = p0 + (p1 - p0) * s + (p2 - p0) * t
s = <0.0,1.0>
t = <0.0,1.0>
s+t<=1.0

While solwing this system:

p.a = p0.a + (p1.a - p0.a) * s + (p2.a - p0.a) * t
p.b = p0.b + (p1.b - p0.b) * s + (p2.b - p0.b) * t
----------------------------------------------------

You got two algebraic options:

I.  t = (p.a - p0.a - (p1.a - p0.a) * s) / (p2.a - p0.a)
II. p.b = p0.b + (p1.b - p0.b) * s + (p2.b - p0.b) * t
----------------------------------------------------
II. p.b = p0.b + (p1.b - p0.b) * s + (p2.b - p0.b) * (p.a - p0.a - (p1.a - p0.a) * s) / (p2.a - p0.a)
II. s = (p.b-p0.b) / ( (p1.b-p0.b) + ( (p2.b-p0.b)*(p.a-p0.a-(p1.a-p0.a)/(p2.a-p0.a) ) )
...

And:

I.  s = (p.a - p0.a - (p2.a - p0.a) * t) / (p1.a - p0.a)
II. p.b = p0.b + (p1.b - p0.b) * s + (p2.b - p0.b) * t
----------------------------------------------------
...

Which gives you 2 algebraic solution options. To ensure stability you should divide with big enough magnitudes. So you should choose the axises (a,b -> x,y) , and point order so you are not dividing by zero or small magnitude numbers.

To avoid this you can use matrix approach

p.a = p0.a + (p1.a - p0.a) * s + (p2.a - p0.a) * t
p.b = p0.b + (p1.b - p0.b) * s + (p2.b - p0.b) * t
--------------------------------------------------
|p.a|   | (p1.a - p0.a) , (p2.a - p0.a) , p0.a |   | s |
|p.b| = | (p1.b - p0.b) , (p2.b - p0.b) , p0.b | * | t |
| 1 |   |       0       ,       0     ,     1  |   | 1 |
--------------------------------------------------------
| s |           | (p1.a - p0.a) , (p2.a - p0.a) , p0.a |   | p.a |
| t | = inverse | (p1.b - p0.b) , (p2.b - p0.b) , p0.b | * | p.b |
| 1 |           |       0       ,       0       ,   1  |   |  1  |
------------------------------------------------------------------

Also here you got more options for axises order, point order so that the inverse matrix is computable. If you use sub-determinant approach for inverse matrix solution then the only thing that should matter is the final division step. So you can choose the orders until you have nonzero determinant.

来源：https://stackoverflow.com/questions/37424174/numerical-stability-of-point-in-triangle-test-with-barycentric-coordinates