How do I explain the Histogram of Oriented Gradients algorithm to a layman?
Finding the "gradient" of a pixel is finding if there is an edge passing through that pixel, the orientation of that egde and how visible is this edge.
As we are taking into account the direction of the edges, we say "oriented gradients".
The "histogram" counts how many pixels have an edge with a specific orientation. The pixels that have visible edges count more than the pixels that have soft edges.
For example if we have a square in the image, we will see that the HOG has a lot of pixels counted in the vertical direction, and the same amount of pixels counted in the horizontal direction, no pixels would get counted in the diagonal directions. If we had a rectangle laying flat, there would be more pixels in the horizontal direction than in the vertical, because the horizontal edges are longer. If we had a diamond, we would count pixels with diagonal edges. This way you can recognize shapes just comparing the histograms (how many pixels have edges in each direction).
If we need to find squares (or faces, or people or anything) of a specific size, we divide the image in blocks of the size of the squares we want to find and compare the HOG that we obtain with the HOG of the thing that we are searching.
I have found the lecture given by Dr. Silvio on HoG to be extremely effective in explaining the concept. I have used to this when I was studying myself and also for explanation to others and it has worked really well every time. HTH
PS - Copyrights for the slides belong to Dr. Silvio and his group.
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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.
I want to fit an image of a clown like face into a contour of another face (a person).
I am detecting the persons face and getting a elliptical-like contour.
I can figure out the center, radius, highest, lowest, left-most and right-most points.
How do I fit the clown face (a square image which I can make elliptical by cutting the face out of the empty background of a png and then detecting the contour) into the persons face?
Or at the least, how do I fit a polygon into another polygon.
I can fit a rectangular image into a rectangular contour with ease, but faces aren't that shape.
Python preferable, but C++ is also manageable, thank you.
Edit: Visual representation as requested:
I have
and I want to make it like this:
but I want the clown face to stretch over the guys face and fit within the blue contour.
I think the keyword you are looking for is Active Appearance Models. First, you need to fit a model to first face (such as this one), which lays inside the contour. Then, you should fit the same model to the clown face. After that, since you have fitted same model to both faces, you can stretch it as you need.
I haven't use AAM myself and I'm not an expert about it, so my explanation might not be enough or might not be exactly correct, but I'm sure it will give you some insight.
A simple and good answer to this question is to find the extreme top, bottom, left, and right points on your contour (head) and then resize your mask to match the aspect ration and place it to cover the 4 points.
Because human heads are elliptical you can use fitEllipse() to give you those 4 points. This will automagically fix any problems with the person tilting their head because regardless of the angle you will know which point is top, bottom, left, and right.
The relevant code for finding the ellipse is:
vector<Point> contour;
// Do whatever you are doing to populate this vector
RotatedRect ellipse = fitEllipse(Mat(contour));
There is also an example as well as documentation for RotatedRect.
// Resize your mask with these sizes for optimum fit
ellipse.size.width
ellipse.size.height
You can rotate your image like this.
UPDATE:
You may also want to find the contour's extreme points to know how much you need to scale your image to ensure that all of the face is covered.
Given two rectangles, and we know the position of four corners, widths, heights, angles.
How to compute the overlapping ratio of these two rectangles?
Can you please help me out?
A convenient way is by the Sutherland-Hodgman polygon clipping algorithm. It works by clipping one of the polygons with the four supporting lines (half-planes) of the other. In the end you get the intersection polygon (at worst an octagon) and find its area by the polygon area formula.
You'll make clipping easier by counter-rotating the polygons around the origin so that one of them becomes axis parallel. This won't change the area.
Note that this approach generalizes easily to two general convex polygons, taking O(N.M) operations. G.T. Toussaint, using the Rotating Caliper principle, reduced the workload to O(N+M), and B. Chazelle & D. P. Dobkin showed that a nonempty intersection can be detected in O(Log(N+M)) operations. This shows that there is probably a little room for improvement for the S-H clipping approach, even though N=M=4 is a tiny problem.
Use rotatedRectangleIntersection function to get contour and use contourArea function to get area and find the ratios
https://docs.opencv.org/3.0-beta/modules/imgproc/doc/structural_analysis_and_shape_descriptors.html#rotatedrectangleintersection
Lets say you have rectangle A and B the you can use the operation:
intersection_area = (A & B).area();
from this area you can calculate de respective ratio towards one of the rectangles. there will be harder more dynamic ways to do this as well.
I have a digital image, and I want to make some calculation based on distances on it. So I need to get the Milimeter/Pixel proportion. What I'm doing right now, is to mark two points wich I know the real world distance, to calculate the Euclidian distance between them, and than obtain the proportion.
The question is, Only with two points can I make the correct Milimeter/Pixel's proportion, or do I need to use 4 points, 2 for the X-Axis and 2 for Y-axis?
If your image is of a flat surface and the camera direction is perpendicular to that surface, then your scale factor should be the same in both directions.
If your image is of a flat surface, but it is tilted relative to the camera, then marking out a rectangle of known proportions on that surface would allow you to compute a perspective transform. (See for example this question)
If your image is of a 3D scene, then of course there is no way in general to convert pixels to distances.
If you know the distance between the points A and B measured on the picture(say in inch) and you also know the number of pixels between the points, you can easily calculate the pixels/inch ratio by dividing <pixels>/<inches>.
I suggest to take the points on the picture such that the line which intersects them is either horizontal either vertical such that calculations do not have errors taking into account the pixels have a rectangular form.
I have to detect the pattern of 6 circles using opencv. I have detected the circles and their centroids by using thresholding and contour function in opencv.
Now I have to define the relation between these circles in a way that should be invariant to scale and rotation. With this I would be able to detect this pattern in various views. I have to use this pattern for determining the object pose.
How can I achieve scale/rotation invariance? Do you have any reference I could read about it?
To make your pattern invariant toward rotation & scale, you have to normalize the direction and the scale when detecting your pattern. Here is a simple algorithm to achieve this
detect centers and circle size (you say you have already achieved this - good!)
compute the average center using a simple mean. Express all the centers from this mean
find the farthest center using a simple norm (euclidian is good enough)
scale the center position and the circle sizes so that this maximum distance is 1.0
rotate the centers so that coordinates of the farthest one is (1.0, 0)
you're done. You are now the proud owner of a scale/rotation invariant pattern detector!! Congratulations!
Now you can find patterns, transform them as suggested, and compare center position & circle sizes.
It is not entirely clear to me if you need to find the rotation, or merely get rid of it, or detect if the circles actually form the pattern you linked. Either way, the answer is much the same.
I would start by finding the two circles that have only one neighbour. For each circle centroid calculate the distance to the closest two neighbours. If the distances differ in more than say 10%, the centroid belongs to an "end" circle (one of the top ones in your link).
Now that you have found the two end circles, rotate them so that they are horizontal to each other. If the other centroids are now above them, rotate another 180 degrees so that the pattern ends up in the orientation you want.
Now you can calculate the scaling from the average inter-centroid distance.
Hope that helps.
Your question sounds exactly like what the SURF algorithm does. It finds groups of interest and groups them together in a way invarant to rotation and scale, and can find the same object in other pictures.
Just search for OpenCV and SURF.