How to get the left chain of points of a polygon? - c++

Question

I am trying to get the left polygonal chain given a set of consecutive points. (NOTE: edges are non-intersecting.)
Image 1. Sample polygon and its bound.
What I did was:
Get the minY, maxY and minX. (Bound.)
Find the point that contains minY (or maxY) then save it as the first point.
Save any points until point with minY or maxY is found while checking for point with minX.
If the same Y is found first, save it as the new first point and repeat from #3.
If other Y is found first and the saved points has minX, this is the chain. Otherwise, save as the new first point and repeat from #3.
Image 2. The left chain of points.
But using this steps might give wrong result for some polygon, like this:
Since one point is (minX, maxY), either of the side will be returned.
EDIT:
With the idea of the left-bottom- and left-top-most points, here is the current code that I am using:
Get the min (left-bottom-most) and max (left-top-most) point.
std::vector<Coord> ret;
size_t i = 0;
Coord minCoord = poly[i];
Coord maxCoord = poly[i];
size_t minIdx = -1;
size_t maxIdx = -1;
size_t cnt = poly.size();
i++;
for (; i < cnt; i++)
{
Coord c = poly[i];
if (c.y < minCoord.y) // new bottom
{
minCoord = c;
minIdx = i;
}
else if (c.y == minCoord.y) // same bottom
{
if (c.x < minCoord.x) // left most
{
minCoord = c;
minIdx = i;
}
}
if (c.y > maxCoord.y) // new top
{
maxCoord = c;
maxIdx = i;
}
else if (c.y == maxCoord.y) // same top
{
if (c.x < maxCoord.x) // left most
{
maxCoord = c;
maxIdx = i;
}
}
}
Get the points connected to the max point.
i = maxIdx;
Coord mid = poly[i];
Coord ray1 = poly[(i + cnt - 1) % cnt];
Coord ray2 = poly[(i + 1) % cnt];
Get which has smallest angle. This will be the path we will follow.
double rad1 = Pts2Rad(mid, ray1);
double rad2 = Pts2Rad(mid, ray2);
int step = 1;
if (rad1 < rad2)
step = cnt - 1;
Save the points.
while (i != minIdx)
{
ret.push_back(poly[i]);
i = (i + step) % cnt;
}
ret.push_back(poly[minIdx]);

Answer 1

To be specific, I am assuming that no vertex is duplicated and define the "left chain" as the sequence of vertices from the original polygon loop that goes from the leftmost vertex in the top side of the bounding box, to the leftmost vertex in the bottom side of the bounding box. [In case the top and bottom sides coincide, these two vertices also coincide; I leave it to you what to return in this case.]
To obtain these, you can scan all vertices and keep the left-topmost so far and left-bottommost so far. Then compare to the next vertex. If above the left-topmost, becomes the new lef-topmost. If at the same level and to the left, becomes the new left-topmost. Similarly for the left-bottommost.