OpenCV find polygons - opencv

I'm using EmguCV and trying to find polygons within an image. Here are some facts about the problem:
1) The polygons are irregularly shaped, but the sides are always at one of two angles.
2) Often the polygons have gaps in their sides that need to be filled.
3) If a polygon is contained within another polygon, I want to ignore it.
Consider this image:
And I want to find the polygons highlighted in red, omit the polygon highlighted in green and make connections across gaps as shown in blue here:
I've had some success using HoughLinesBinary and then connecting the closest line segment end points to each other to bridge gaps to build a complete polygon, but this doesn't work when multiple polygons are involved since it will try to draw lines between polygons if they happen to be close to each other.
Anybody have any ideas?

I think that the problem could be your image threshold.
I don't know how you did but you can get better results if the binary image is better.
I like of your idea to connect the polygons segment, try to secure that the line you will connect has a maximum and minimum length to avoid connect to nearest objects.
Verify if the new joint of lines forms 90 degrees angle, even if its a corner.
Added:
You can use moephological operators to grow the lines acord to angles. As you said that the lines has known angles do dilations using a mask like this
0 0 0
1 1 0
0 0 0
This mask will grow lines only to rigth until conect to other side.

A general solution will be hard, but for your particular problem a relatively simple heuristic should work.
The main idea is to use the white pixels on one side of a wall as an additional feature. The walls in your image always have one almost black side, and one side with many white noise pixels. The orientation of the white noise in relation to the wall does not switch on corners, so using this information you can eliminate a lot of possible connections between lines.
First some definitions:
All walls in the picture go either from the lower left to the upper right (a rising line) or from the upper left to the lower right (a falling line).
If there are more white pixels on the left side of a falling line, call it falling-left-wall, otherwise falling-right-wall. Same for rising lines.
Each line ends in two points. Call the leftmost one start, the rightmost one end.
Now for the algorithm:
classify each line and each start/endpoint in the image.
Check the immediate area on both sides of each line, and see which side contains more white pixels.
Afterwards, you have a list of points labeled falling-left-wall-start, rising-left-wall-end, etc.
for each start/end point:
look for nearby start/end points from another line
if it is a falling-left-wall-start, only look for:
a falling-left-wall-end
a rising-left-wall-start, if that point is on the left side of the current line
a rising-right-wall-end, if that point is on the right side of the current line
pick the closest point among the found points, connect it to the current point.

Related

Finding point height on a cup using OpenCV

Suppose that I want to find the 3D position of a cup with its rotation, with image input like this (this cup can be rotated to point in any direction):
Given that I have a bunch of 2D points specifying the top circle and bottom circle like the following image. (Let's assume that these points are given by a person drawing the lines around the cup, so it won't be very accurate. Ellipse fitting or SolvePnP might be needed to recover a good approximation. And the bottom circle is not a complete circle, it's just part of a circle. Sometimes the top part will be occluded as well so we cannot rely that there will be a complete circle)
I also know the physical radius of the top and bottom circle, and the distance between them by using a ruler to measure them beforehand.
I want to find the complete 2 circle like following image (I think I need to find the position of the cup and its up direction before I could project the complete circles):
Let's say that my ultimate goal is to be able to find the closest 2D top point and closest 2D bottom point, given a 2D point on the side of the cup, like the following image:
A point can also be inside of the cup, like so:
Let's define distance(a, b) as a function that find euclidean distance from point a and point b in pixel units.
From that I would be able to calculate the distance(side point, bottom point) / distance(top point, bottom point) which will be a scale number from 0 to 1, if I multiply this number to the physical height of the cup measured by the ruler, then I will know how high the point is from the bottom of the cup in metric unit.
What is the method I can use to find the corresponding top and bottom point given point on the side, so that I can finally find out the height of the point from the bottom of the cup?
I'm thinking of using PnP to solve this but my points do not have correct IDs associated with them. And I don't want to know the exact rotation of the cup, I only want to know the up direction of the cup.
I also think that fitting the ellipse might help somewhat, but maybe it's not the best because the circle is not complete.
If you have any suggestions, please tell me how to obtain the point height from the bottom of the cup.
Given the accuracy issues, I don't think it is worth performing a 3D reconstruction of the cone.
I would perform a "standard" ellipse fit on the top outline, which is the most accurate, then a constrained one on the bottom, knowing the position of the vertical axis. After reduction of the coordinates, the bottom ellipse can be written as
x²/a² + (y - h)²/b² = 1
which can be solved by least-squares.
Note that it could be advantageous to ask the user to point at the endpoints of the straight edges at the bottom, plus the lowest point, instead of the whole curve.
Solving for the closest top and bottom points is a pure 2D problem (draw the line through the given point and the intersection of the sides, and find the intersection points with the ellipse.

Calculation of center point for the localization of robot in 3D data

I am trying to find a reliable method to calculate the corner points of a container. From these corner point’s idea is to calculate the center point of the container for the localization of robot, it means that the calculated center point will be the destination of robot in order to pick the container. For this I am looking for any suggestions to calculate the corner points or may be if any possibility to calculate the center point directly. Up to this point PCL library C/C++ is used for the processing of the 3D data.
The image below is the screenshot of the container.
thanks in advance.
afterApplyingPassthrough
I did the following things:
I binarized the image (black pixels = 0, green pixels = 1),
inverted the image (black pixels = 1, green pixels = 0),
eroded the image with 3x3 kernel N-times and dilated it with same kernel M-times.
Left: N=2, M=1;Right: N=6, M=6
After that:
I computed contours of all non-zero areas and
removed the contour that surrounded entire image.
This are the contours that remained:
I do not know how "typical" input image looks like in your case. Since I only have access to one sample image, I would rather not speculate about "general solution" that will be suitable for you. But to solve this particular case, you could analyze every contour in the following way:
compute rotatated rectangle that fits best around your contour (you need something similar to minAreaRect from OpenCV)
compute areas of rectangle and contour interior
if the difference between contour area and the area of the rotated bounding rectangle is small, the contour has approximately rectangular shape
find the contour that is both rectangular and satisfies some other condition (for example: typical area of the container). Assume that this belongs to container and compute its center.
I am not claiming that this is a solution that will work well in real world scenarios. It is also not fast. You should view it as a "sketch" that shows how to extract some useful information.
I assume the wheels maintain the cart a known offset from the floor and you can identify the floor. Filter out all points which are too close to the floor (this will remove wheels and everything but cart which will help limit data and simplify later steps.
If you isolate the cart, you could apply a simple average point (centroid), alternately, if that is not precise, you could try finding the bounding box of the isolated cart (min max in primary directions) and then take the centroid of that bounding box (this should be more accurate, but will still need a slight vertical offset due to the top handles).
If you can not isolate the cart or the other methods are not working well, you could try using PCL sample consensus specifically SACMODEL_LINE. This will be an involved strategy, but will give very solid results, basically run through and find each line and subtract its members from the cloud so as to find the next best line. After you have your 4 primary cart lines, use their parameters to find your centroid. *this would also be robust against random items being in or on the cart as well as carts of various sizes (assuming they always had linear perpendicular walls)

Opencv divide a contour in two sections

I have a contour in Opencv with a convexity defect (the one in red) and I want to cut that contour in two parts, horizontally traversing that point, is there anyway to do it, so I just get the contour marked in yellow?
Image describing the problem
That's an interesting question. There are some solutions based on how the concavity points are distributed in your image.
1) If such points does not occur at the bottom of the contour (like your simple example). Then here is a pseudo-code.
Find convex hull C of the image I.
Subtract I from C, that will give you the concavity areas (like the black triangle between two white triangles in your example).
The point with the minimum y value in that area gives you the horizontal line to cut.
2) If such points can occur anywhere, you need a more intelligent algorithm which has cut lines that are not constrained by only being horizontal (because the min-y point of that difference will be the min-y of the image). You can find the "inner-most" corner points, and connect them to each other. You can recursively cut the remainder in y-,x+,y+,x- directions. It really depends on the specs of your input.

Finding All Pixels Within Certain Range in Polar Coordinates

I want to find all pixels in an image (in Cartesian coordinates) which lie within certain polar range, r_min r_max theta_min and theta_max. So in other words I have some annular section defined with the parameters mentioned above and I want to find integer x,y coordinates of the pixels which lie within it. The brute force solution comes to mid offcourse (going through all the pixels of the image and checking if it is within it) but I am wondering if there is some more efficient solution to it.
Thanks
In the brute force solution, you can first determine the tight bounding box of the area, by computing the four vertexes and including the four cardinal extreme points as needed. Then for every pixel, you will have to evaluate two circles (quadratic expressions) and two straight lines (linear expressions). By doing the computation incrementally (X => X+1) the number of operations drops to about nothing.
Inside a circle
f(X,Y) = X²+Y²-2XXc-2YYc+Xc²+Yc²-R² <= 0
Incrementally,
f(X+1,Y) = f(X,Y)+2X+1-2Xc <= 0
If you really want to avoid that overhead, you will resort to scanline conversion techniques. First think of filling a slanted rectangle. Drawing two horizontal lines by the intermediate vertices, you decompose the rectangle in two triangles and a parallelogram. Then for any scanline that crosses one of these shapes, you know beforehand what pair of sides you will intersect. From there, you know what portion of the scanline you need to fill.
You can generalize to any shape, in particular your circle segment. Be prepared to a relatively subtle case analysis, but finding the intersections themselves isn't so hard. It may help to split the domain with a vertical through the center so that any horizontal always meets the outline twice, never four times.
We'll assume the center of the section is at 0,0 for simplicity. If not, it's easy to change by offsetting all the coordinates.
For each possible y coordinate from r_max to -r_max, find the x coordinates of the circle of both radii: -sqrt(r*r-y*y) and sqrt(r*r-y*y). For every point that is inside the r_max circle and outside the r_min circle, it might be part of the section and will need further testing.
Now do the same x coordinate calculations, but this time with the line segments described by the angles. You'll need some conditional logic to determine which side of the line is inside and which is outside, and whether it affects the upper or lower part of the section.

OpenCV parallel lines cull

I have an algorithm that simply goes through a number of corners and finds those which are parallel. My problem is, as shown below, that I sometimes get false positive results.
To eliminate this I was going to check if both points fell onto a single hough line but this would be quite computationally intensive and I was wondering if anyone had any simpler ideas.
Thanks.
Ok based on the comments, this should be fix-able. When you detect a pair of parallel lines, get the equation of the line using the two corners that you've used to construct it. This line may be of the form y = mx + c. Then for every y coordinate between the two points, compute the x coordinate. This gives you a set of all the pixels that the line segment covers. Go through these pixels, and check if the intensity at every pixel is closer to black than white. If a majority of the pixels in the set are black-ish, then it's a line. If not, it's probably a non-line.

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