mgs/quickhull
2
0
Fork 0
Code Issues Pull Requests Projects Releases 1 Wiki Activity

Fix rectangle behavior properly, minor improvements

Browse Source
This commit is contained in:
karl 2020-12-06 19:37:47 +01:00
parent 14e633d236
commit 82c0df1840
2 changed files with 9 additions and 16 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

3
Line.h
View File

@ -42,7 +42,8 @@ public:
other -= from();
// Cross product greater than zero?
if (m_to_from_origin.x() * other.y() - m_to_from_origin.y() * other.x() > 0)
// Note: We need ">=" here so that we include points which are directly on the existing hull.
if (m_to_from_origin.x() * other.y() - m_to_from_origin.y() * other.x() >= 0)
{
return true;
}

22
Quickhull.h
View File

@ -132,24 +132,20 @@ private:
// these properly? We definitely need to remove them all from input later;
// do we also need to handle them all further? This hotfix works, but seems
// like an unnecessarily big performance hit for that edge case.
// FIXME: This workaround also causes problems with extremely large numbers
// of randomly generated numbers, causing random lines within the data!
// Points inside the hull are added because of random lines.
for (const Point &point : input)
{
float this_distance = line.distance_squared_to(point);
// TODO: Both are required, otherwise only one side of the rectangle is
// taken -- why?
if (this_distance == furthest_distance || line.distance_squared_to(point) == 0)
if (this_distance == furthest_distance)
{
output.emplace_back(point);
}
}
// Remove the previously checked points from the input -- we can't do that in
// the previous loop because we can't delete from the list we're iterating over
input.remove_if([furthest_distance, line](Point point)
{
return furthest_distance == line.distance_squared_to(point);
});
// Hotfix end
output.emplace_back(furthest_point);
input.remove(furthest_point);
@ -173,12 +169,9 @@ private:
new_line2 = Line(furthest_point, line.to());
}
// TODO: Test if this improves performance
//Triangle triangle(a, b, c);
//input.remove_if([triangle](Point point)
//{
// return triangle.is_point_inside(point);
//});
// We don't need to remove points inside the triangle created by those lines.
// That happens implicitly since they are not handled further due to the
// following step, which assigns each line its corresponding points to handle:
// Get points right of new_line1 and 2
std::list<Point> left_of_line1, left_of_line2;
@ -189,8 +182,7 @@ private:
{
left_of_line1.emplace_back(point);
}
// TODO: Can we do else if here, or could we then miss out on points?
if (!new_line2.is_point_right(point))
else if (!new_line2.is_point_right(point))
{
left_of_line2.emplace_back(point);
}