Now our equation becomes ax+by+c=0 â w1x1+w2x2+w0=0. In Machine Learning we maximum time deals with the nD array so there is no such thing as x,y there we represent as (x1,x2) and constant a,b,c in the form of W. 2] C is different for every line if it becomes the same then θ will be different. We need to keep two things in mind 1] Everywhere θ is the same for a unique line. We can write like this ax+by+c=0 â so from this equation we can say that a/b=m=slope of line. y=mx+C where Y and X are the coordinates, m is the slope and C is the constant distance from the origin. Line (2D)Īs we are studying from our Highschool about line and derivation of line. So, We come to the point Line(2D),Plane(3D),Hyperplane(nD). You cannot measure the length of a ray.Īctually, before reading about âLinesâ you should be clear about line, ray, and segment differences. RayĪ ray is a part of a line that has one endpoint and goes on infinitely in only one direction. You can measure the length of a segment, but not of a line. It contains these endpoints and all the points of the line between them. Line SegmentĪ line segment has two endpoints. When we draw lines in geometry, we use an arrow at each end to show that it extends infinitely. Roughly, an infinitely long collection of points extending in two opposite directions. It is actually difficult to give a good mathematical definition. ![]() I think the above should be enough to get you started and if anything is missing, please read 3 carefully and see corresponding implementations in 2.Line (2D), Plane (3D) and Hyperplane (nD) LINE determine the distance to the closest point in the subspaceÄist=calculateDistance(imageOne, imageTwo, (const double **) tangents, numTangentsRemaining, height, width) NumTangentsRemaining = normalizeTangents(tangents, numTangents, height, width) Just follow the implementation in tangentDistance function: // determine the tangents of the first imageĬalculateTangents(imageOne, tangents, numTangents, height, width, choice, background) I like the below image from 3, which clearly shows the displacements effect of different transformation tangents.Ĭalculating the tangent distance between images You can find all other tangents described in 3 implemented at 2. So you know that is exactly the formula used in 3. Similarly - (j+offsetW)*factor corresponds to -x. Using the conclusion from previous, we know (k+offsetH)*factor corresponds to y. Rotation tangent is defined as below in 3:ĬurrentTangent = ((k+offsetH)*x1 - (j+offsetW)*x2)*factor ![]() Similarly you know the vertical displacement (k+offsetH)*factor. So I guess (j+offsetW)*factor is the displacement for each pixel, which is proportional to the horizontal distance from the pixel to the center of the image. Since j(also k) is an integer between 0 and 15 inclusive (the width and the height of the image are 16 pixels), (j-7.5)/8 is just a fraction number between -0.9375 to 0.9375. So that, we plug in those constants: currentTangent = ((j-7.5)*x1 + (k-7.5)*x2)/8 The author is using images with size 16x16. In the beginning of td.c in 2's implementation, we know the below definition: factorW=((double)width*0.5) įactor=(factorH ![]() ![]() From 3, we know the transformation is as below:ĬurrentTangent = ((j+offsetW)*x1 + (k+offsetH)*x2)*factor Here I show you one of the tangent calculations implemented in 2 - the scaling tangent. If you need more background about image convolution, here is an example.Īfter this step is done, you have calculated the horizontal and vertical shift: The author suggests to use a Gaussian function. This step is to map the discrete vector to continuous one so that it is differentiable. Below we show the result of 3 different smooth operations (check section 4 of 3) (The left column shows the result images, the right column shows the original images and the convolution operators).
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