Re: [RFC] Gradient fill options

[Daniel J S. <dan...@ie...>]

Hans-Bernhard Bröker wrote:
> Daniel J Sebald wrote:
> 
>>I've looked through the code a bit and your examples.  (What command 
>>does one need to compile this program?)  I'd have to think about your 
>>approach to be certain about the result.  You've broken the thing down 
>>into a series of one dimensional linearities, first along the sides of 
>>the triangle to get a color, and then across the x dimension to further 
>>interpolate between those two.  Seems logical, but it may be open for 
>>rounding problems.
> 
> 
> And, more importantly, it's a bit biased, which tends to turn into a 
> robustness problem.  The bias lies in the arbitrary choice of one vertex 
> to be shared by the two first interpolations.
> 
> It can be better to go via barycentric coordinates.  It's effectively a 
> method to find three numbers s, t, and u such that:
> 
> 	s+t+u=1
> 	s*A + t*B + u*C = X  for the point, and its colour.
> 
> I.e. you solve the equation for s,t,u given a point X, then substitute 
> them into the equation to find the target colour.

Interesting.  This results in matrix equations similar to what I sent previously:

s * x_1 + t * x_2 + u * x_3 = x
s * y_1 + t * y_2 + u * x_3 = y
s * 1   + t * 1   + u * 1   = 1

or

| x_1 x_2 x_3 |
| y_1 y_2 y_3 |
| 1   1   1   |

(call it Z) product with

[s t u]'

equals

[x y 1]'

so

[s t u]' = inv(Z) [x y 1]'

Since I doubt you'd want to compute the intermediate values s t u for each pixel inside the triangle

c = [c_1 c_2 c_3] * [s t u]'

 = [c_1 c_2 c_3] * inv(Z) * [x y 1]'

 = [w_1 w_2 b] * [x y 1]'

 = w' * [x y]' + b

Dan