What is the fastest way of converting a quadrilateral (made up of four x,y
points) to a triangle strip? I'm well aware of the general triangulation algorithms that exist, but I need a short, well optimized algorithm that deals with quadrilaterals only.
My current algorithm does this, which works for most quads but still gets the points mixed up for some:
#define fp(f) bounds.p##f
/* Sort four points in ascending order by their Y values */
point_sort4_y(&fp(1), &fp(2), &fp(3), &fp(4));
/* Bottom two */
if (fminf(-fp(1).x, -fp(2).x) == -fp(2).x)
{
out_quad.p1 = fp(2);
out_quad.p2 = fp(1);
}
else
{
out_quad.p1 = fp(1);
out_quad.p2 = fp(2);
}
/* Top two */
if (fminf(-fp(3).x, -fp(4).x) == -fp(3).x)
{
out_quad.p3 = fp(3);
out_quad.p4 = fp(4);
}
else
{
out_quad.p3 = fp(4);
out_quad.p4 = fp(3);
}
Edit: I'm asking about converting a single quad to a single triangle strip that should consist of four points.
Given a quad ABCD
we can split it into ABC, ACD
or ABD, DBC
.
Compare the length of AC
and BD
and use the shorter for the splitting edge. In other words use ABC, ACD
if AC
is shorter and ABD, DBC
otherwise.
p
, where p
is an iterator on a cyclic container p + 2
lies inside of the ear formed by {p - 1, p, p + 1}
. If yes, seek extremum wrt other coordinate (or switch from min
to max
or vice versa) and repeat step 1. t0 = { p - 1, p, p + 1}
and t1 = { p + 1, p + 2, p - 1 }
No need to sort, just locate the extremum. If your quads are guaranteed to be convex (real quadrilaterals) then skip the extremum search and choose arbitrary p
.
Edit: Modified according to suggestion from the comment. Also the formulation suggested by the commenter is simpler to implement:
A, B, C, D
form the diagonals AC
and BD
B
and D
lie on different sides of AC
then AC
can be used to split the quad BD
and points A
and C
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