A library can be used to capture the image from the screen. https://pypi.python.org/pypi/pyscreenshot ♪ The library works for all operating systems different from PIL.ImageGrab. We need to use the method. grab♪ This method, as an input parameter, accepts the coordinates of the desired area and then returns the image. To transform the image into a numpy, it's enough to use the challenge. numpy.array♪That's how you can seize three sites and transform them into a numpy mass:import pyscreenshot as ps import numpy as np boxes = [(0, 0, 100, 100), (100, 100, 150, 150), (300, 300, 366, 350)] arrays = [] for i in range(3): im = np.array(ps.grab(bbox=boxes[i])) arrays.append(np.array(im)) for array in arrays: print("Type:", type(array)) print("dtype:", array.dtype) print("shape:", array.shape) print()

Personally, I use Docker Quickstart Terminal, who's on delivery. He's going to get a linux virtual, and he's gonna have to dig up the image of the openfice and start it. And then you feed your photos before you cook them.docker pull bamos/openface docker run -p 9000:9000 -p 8000:8000 -t -i bamos/openface /bin/bash

#include <opencv2/opencv.hpp> using namespace cv; int main(int, char**) { VideoCapture cap(0); // открываем камеру if(!cap.isOpened()) // Mat edges; namedWindow("edges",1); for(;;) { Mat frame; cap >> frame; // получаем кадры от камеры cvtColor(frame, edges, COLOR_BGR2GRAY); GaussianBlur(edges, edges, Size(7,7), 1.5, 1.5); Canny(edges, edges, 0, 30, 3); imshow("edges", edges); if(waitKey(30) >= 0) break; } // деструктор return 0; }

The answer is still found. The fault was not detected during the last de facto line: imgGray.Dispose();imageRotate = imgGray.Rotate(-avg, new Gray(255)); } else { imageRotate = imgGray; } imgGray.Dispose(); //return imageRotate.Bitmap; return BlackFilterImage(imageRotate); } private static Bitmap BlackFilterImage(Image&lt;Gray, byte&gt; img) { UMat resultImg = new UMat(); CvInvoke.Threshold(img, resultImg, 105, 255, ThresholdType.BinaryInv); //Ошибка возникала тут Unfortunately, the nature of the error led me to a false trail.

You need to retrieve these instructions: Installation of CMake, first of all, the DMG-Fail of the binary distributive https://cmake.org/download/ Once the installation is completed, you will be I'm asking you to bleed CMake into /usr/bin Choose yes and finish. Installation to verify that CMake, Put it in Terminal. CMake -versionOpenCV installation First, download the original files of the stable version. https://sourceforge.net/projects/opencvlibrary/files/ (in my case 2.4.7) Pick up the files in the folder and go to that folder through the terminal, for example.cd ~/Downloads tar -xvf opencv-2.4.7.tar cd opencv-2.4.7Now, we're going to collect and set up the OpenCV in Terminal, the next teams have to be successfully completed (without mistakes)mkdir build cd build cmake -G "Unix Makefiles." make -j8 sudo make install

Copy opencv***.whl in the folder...python35/Scripts and set cv2 (pip install opencv**.whl). Anaconda and idle are different programs. Checking the packages, team pip list, in the Scripts file, through the console.

I realized the mistake. It's not about parameters, it's that I tried to use a vector from another method when, as a first-stage developer, I cleaned that vector, and calcOpticalFlowPyrLK I read two empty originators, the one that was supposed to be with points, and the one with pyramids. We had to use a dynamic vector.The pyramids should've built a pyramid, not an object, but a footage where the object was found.

OpenCV provides many tools to recognize samples, including through contour analysis. In the third version of the Fremworth, almost everything with regard to the comparison of contours has been assigned to a separate module with a title http://docs.opencv.org/master/d1/d85/group__shape.html ♪The author of the question did not specify which figures (bands of figures) would be compared to the infinity sign, and only noted that interest was only the fact that the was likely to be a candidate.Obviously, the notion of "finality" in itself would not say anything, but it would therefore have to offer her some form of reference. Let this image look like that. model1.jpgHowever, in order to make sure that the method further revealed allows for different features, add several different images (Files) to the reference model. model2.jpg♪ model3.jpg and model4.jpg respectively):In fact, all images of the same size and are shown in a reduced form only to save space on the page.Let's start by deleting the object the author gave the question:cv::Mat tst_src_mat = cv::imread("test.jpg", cv::IMREAD_GRAYSCALE); if(tst_src_mat.empty()) return; cv::Mat tst_bin_mat; cv::threshold(tst_src_mat, tst_bin_mat, 0, 255 , cv::THRESH_OTSU | cv::THRESH_BINARY_INV); After the binarization, there will be an artifact in addition to the object of interest in the image in the left bottom corner:It could be eliminated through a simple and philosophical approach, without paying attention, leaving only the contours of interest in the contrary. This can be done, for example, by focusing on the maximum area and the corresponding index in the vector:std::vector<std::vector<cv::Point> > tst_cs; cv::findContours(tst_bin_mat.clone(), tst_cs , cv::RETR_EXTERNAL, cv::CHAIN_APPROX_SIMPLE); int ci = -1; double max_area = 0.; for(int i = 0, n = tst_cs.size(); i < n; ++i) { const double area = cv::contourArea(tst_cs.at(i)); if(area > max_area) {max_area = area; ci = i;} } if(ci == -1) return; It follows a nuance which implies that any comparable contours shall have the same number of points in their composition. Of course, the different figures will not normally follow this rule, which means they're gonna have to help it by adding the required number of points, but only at the coordinates that already exist in the figure:const int num_pts = 300; for(int i = tst_cs.at(ci).size()-1, d = 0; i < num_pts; ++i) tst_cs[ci].push_back(tst_cs[ci][d++]); Constant num_pts - this is an arbitrary number. It depends, in principle, on the complexity of the Ideally, it should be equal to the maximum number of points at the e-e-e... a multi-point contour the compared.Perhaps it is time to turn to the previously considered reference and images that are very different from the test figure:// Герой настоящей статьи, который собственно // и будет производить сравнение контуров. cv::Ptr<cv::ShapeContextDistanceExtractor> sc = cv::createShapeContextDistanceExtractor(); for(int i = 0; i < 4; ++i) { std::string fname = std::string("model") + std::to_string(i+1) + std::string(".jpg"); cv::Mat src_mat = cv::imread(fname, cv::IMREAD_GRAYSCALE); if(src_mat.empty()) return -1; cv::Mat bin_mat; cv::threshold(src_mat, bin_mat, 0, 255 , cv::THRESH_OTSU | cv::THRESH_BINARY_INV); std::vector&lt;std::vector&lt;cv::Point&gt; &gt; cs; cv::findContours(bin_mat.clone(), cs , cv::RETR_EXTERNAL, cv::CHAIN_APPROX_SIMPLE); for(int i = cs.at(0).size()-1, d = 0; i &lt; num_pts; ++i) cs[0].push_back(cs[0][d++]); std::cout &lt;&lt; fname &lt;&lt; " - " &lt;&lt; sc-&gt;computeDistance(tst_cs[ci], cs[0]) &lt;&lt; std::endl; } The result of the code will be such:model1.jpg - 2.84489model2.jpg - 112.033model3.jpg - 32.8308model4.jpg - 524.659Presentation model1.jpg contains infinity, and consequently has the lowest distance between the test figure. Sobca images.model2.jpg) and hammer serpa.model4.jpg(d) In view of the obvious differences, the list of irregularities has been completed. But who would have thought "Betman"model3.jpgThis is the second place. But if you look around, it's quite possible to catch a common with an endless line.

RasPI contains an excellent MIPI CSI interface for cameras, state chambers (acquired separately) allow the use of fullHD/30, HD/60 VGA/90. The RasPI processor may not be enough. FullHD won't last, and you certainly don't need FullHD. Take an extra long-focused object for RasPI camera. Amazon of their pile, focus on a certain range where The number of passing cars is the most visible. Take the VGA video.for recognition can be used http://www.openalpr.com/ RasPI will beOf course, to keep up because Tesseract is using. But they sayIt's patient.You can. https://en.wikipedia.org/wiki/Comparison_of_single-board_computers It's not like small, but it's mandatory for MIPI CSI CSI and that the cameras are ready for service. Good video- https://www.youtube.com/watch?v=oBAOdj8HABc https://www.youtube.com/watch?v=ZcMPemzifbQ

The time spent on finding a solution had shown that it was difficult to disagree with the authors of comments on the use of washing and clustering. It was found that the second paragraph (in fact clustering) could be used by K-Med. Of course, provided that the number of colour strips on the image is known in advance:const int cluster_count = 10; Loading the image from the file and breaking its matrix into separate channels:cv::Mat src_mat = cv::imread("image.jpg"); if(src_mat.empty()) return; std::vector<cv::Mat> channels; cv::split(src_mat, channels); By default, OpenCV will download the picture. BGR format. It can be converted into other coloured areas (e.g. HSV or YCbCrbut there has been no further benefit. I've even been trying to combine different channels of different colour spaces among themselves, but this has also not improved.It is further necessary to establish a matrix with a height equal to the number of points of reference and the width to the number of channels previously obtained:const int rows = src_mat.rows * src_mat.cols; const int cols = channels.size(); cv::Mat mat = cv::Mat::zeros(rows, cols, CV_8U); for(int col = 0; col < cols; ++col) { // Применяем размытие с ядром 4х4. // Как оказалось, это даёт самый лучший результат. // Больший размер ядра фильтра размытия ведёт // к большей деформации результата, так что 4х4 - это компромисс. cv::Mat blr_mat; cv::blur(channels[col], blr_mat, cv::Size(4,4)); // Размытое изображение переформатируем в матрицу с одной колонкой. blr_mat.reshape(1,rows).copyTo(mat.col(col)); } Convergate the matrices in the type of number with float:mat.convertTo(mat, CV_32F); There's method of clustering K-medium:cv::Mat labels_mat; cv::kmeans(mat, cluster_count, labels_mat, cv::TermCriteria(), 1 , cv::KMEANS_PP_CENTERS); Now. labels_mat will contain the values of each picsel of a cluster. For ease of drawing, these values may be normalized:cv::convertScaleAbs(labels_mat.reshape(0,src_mat.rows) , labels_mat, 255/cluster_count, 1); Outcome: It is not clear that everything is far from perfect, but in general, the flower strips are correct and simply need to work on the elimination of the artifacts. This can already be done through morphological operations, or even through contour processing, dislocating small in their area.

The simplest way is to calculate the difference between RGB samples and the test color. About:// класс цвета class color { public: string _name; char _r, _g, _b; // конструктор color (string name, char r, char g, char b) { _name = name; _r = r; _g = g; _b = b; } // сравнение с другим цветом int distanceTo (color &amp;another) { return (abs(another._r - _r) + abs(another._g - _g) + abs(another._b - _b)); } } ... // набор цветов для сравнения vector<color> sources; sources.push_back(color("red", 255, 0, 0)); sources.push_back(color("green", 0, 255, 0)); sources.push_back(color("blue", 0, 0, 255)); sources.push_back(color("yellow", 255, 255, 0)); sources.push_back(color("purple", 255, 0, 255)); sources.push_back(color("black", 0, 0, 0)); sources.push_back(color("white", 255, 255, 255)); // интересующий цвет color test("test", 64, 32, 32); // лучшее совпадение color *best = nullptr; int dst = 1024; // поиск for (auto it = sources.begin; it != sources.end(); it++) { int current = (*it).distanceTo(test); if (current < dst) { dst = current; best = *(*it); } } // вывод ближайшего совпадения cout << "Best match: " << best->name << "\n"; A more precise result can be obtained by comparison in colours https://ru.wikipedia.org/wiki/HSV_%28%D1%86%D0%B2%D0%B5%D1%82%D0%BE%D0%B2%D0%B0%D1%8F_%D0%BC%D0%BE%D0%B4%D0%B5%D0%BB%D1%8C%29 ♪ But in most cases, there are also enough comparisons channels.

I knew what the hell it was. When I had VS2013it worked well. But then he broke, he didn't want to recover, and I started a project. VS2015♪ Thinking that algorithm that worked well before, and now it's doing great, it never looked. When I looked, sword. Okay, now I've moved all my code to a clean project, and it's working. Thank you all.

Regrettably, the code (see discussion in the commentaries on the question) does not show where there may be a mistake. Nevertheless, the minimum example that reproduces a problem must look somewhat different. Like that.class CaptureWrapper { public: CaptureWrapper() {} ~CaptureWrapper() {} bool open() { return _vid_cap.open(0); } void close() { if(_vid_cap.isOpened()) _vid_cap.release(); } private: cv::VideoCapture _vid_cap; }; int main(int argc, char *argv[]) { CaptureWrapper wrapper; if(wrapper.open() == false) { std::cout << "Failed!"; return -1; } wrapper.close(); return 0; } If the minimum example is in place, its function may be gradually increased, such as the possibility of selecting an index device or the way to the video file, each checking the correctness of the application. So it's gonna be easy to locate a troubled place in the code if it happens.

Throwsthreshold) is the arbitrary values used by the Kenny algorithm to decide on each of the peaks at the final stage of its work. The thresholds are two: upper and lower.If the magnetic value of each individual picel is greater than the upper threshold, the picsel is considered to be a border. If the magnetic value of the gradient is less than the lower threshold, the picsel shall be removed (no longer considered).If the value of the picsel gradient gradient is between the thresholds, in this situation, the algorithms are tested either separately in different implementations, based on the values and directions of the gradient at the adjacent peaks or, as is the case in the OpenCV, mark it as a border, provided that the pics in question are adjacent to the picsel of the radiant top.Dimension of the kernelkernel size or aperture size) is the size of the square brick matrix (convolutionwhich are used to calculate the gradient on the vertical and horizontal directions. In the case of OpenCV, the permissible size of the kernels can only be "3", "5" or "7." This function is performed by the operator of the Area, who is transparently called into the function Canny()♪The imaging curve is essentially a new computation operation for each picsel based on the values of the other piccells that are his neighbours. In addition to the image itself, a separate matrix, often called a window or filter, is involved. Usually use a square one with equals, but with an odd number of lines and columns. Incorrectness is needed to obtain the central element of the matrix or otherwise the so-called anchor:( )( )( ) ( )(x)( ) ( )( )( ) Every picsel of the reference image shall be stacked by an anchor, thus closing the filter and all adjacent to the current peak. The sum of the peaks on the filter elements is then calculated and will be the new peak value.

It's better to start right away from the second paragraph, because I can assume it's a priority.Bionarization (steering) is essentially a process whereby the amount of information available on the image is minimized. This fact is both a plus and a minus, since, together with uninterestested information, information is erased and that could have greatly contributed to the task.It is not difficult to see the presence of white ozone on the right side. These "zones" are of the same size and, in general, they are not separated from the exact same white "zone" but are already in the field of interest (parks).White &quot; zones &quot; within the area of interest may also have an arbitrary size and often cannot be considered as a whole with others in the vicinity, as they have separate boundaries and relatively close size with those that are not in the area of interest.By all means, the "zones" between the two areas of interest can be closer to each other (by the distance between the moments of contour) than the "zones" of the same parking spot.All of the above-mentioned exceptions are fully presented in the picture of the photoshop, which actually makes it impossible to correctly cluster the ozone contour into the automobile.By binarization, an acceptable result is extremely rare if the original image contains a natural environment.However, if a person is allowed to interfere with the algorithm, namely, to indicate to the last real boundaries of the facility by hand-delivering parking rectangles, the whole task will be a fairly simple operation. Example of binarization of the C++ and C# (EmguCV).Suppose there is a reference: We'll load it, translate it into shades of grey and binarize:C+++:cv::Mat src_mat = cv::imread("img.jpg"); if(src_mat.empty()) return; cv::Mat gry_mat; cv::cvtColor(src_mat, gry_mat, cv::COLOR_BGR2GRAY); cv::Mat bin1_mat, bin2_mat; cv::threshold(gry_mat, bin1_mat, 230, 255, cv::THRESH_BINARY); cv::threshold(gry_mat, bin2_mat, 25, 255, cv::THRESH_BINARY_INV); cv::Mat bin_mat = bin1_mat + bin2_mat; C#:Mat srcMat = new Mat("img.jpg", LoadImageType.AnyColor); if(srcMat.IsEmpty) return; Mat grayMat = new Mat(); Mat bin1Mat = new Mat(); Mat bin2Mat = new Mat(); Mat binMat = new Mat(); CvInvoke.CvtColor(srcMat, grayMat, ColorConversion.Bgr2Gray); CvInvoke.Threshold(grayMat, bin1Mat, 230, 255, ThresholdType.Binary); CvInvoke.Threshold(grayMat, bin2Mat, 25, 255, ThresholdType.BinaryInv); CvInvoke.Add(bin1Mat, bin2Mat, binMat); Here, an attempt has been made to binarize the upper and lower borders. The coefficients are selected manually and may differ as best for other images. Thus, more useful details are retained: Next, if we start looking for contours and paint them:C+++:std::vector<std::vector<cv::Point> > contours; cv::findContours(bin_mat.clone(), contours , cv::RETR_EXTERNAL, cv::CHAIN_APPROX_SIMPLE); cv::Mat ctr_mat = cv::Mat::zeros(bin_mat.size(), CV_8U); cv::drawContours(ctr_mat, contours, -1, cv::Scalar::all(255), 1); C#:VectorOfVectorOfPoint contours = new VectorOfVectorOfPoint(); Mat hierarchy; CvInvoke.FindContours(binMat, contours, hierarchy, RetrType.External, ChainApproxMethod.ChainApproxSimple); Mat ctrMat = new Mat(binMat.Size, DepthType.Cv8U, 3); CvInvoke.DrawContours(ctrMat, contours, -1, new MCvScalar(255)); ...we'll get a picture like this: Contours have proved to be quite large, and there is already doubt that the contours of individual parts can be assembled in a single contour of the vehicle. Let's try to filter some of them, for example, in the square, while drawing them on the specimen of quadratics in the original image:C+++:std::vector<std::vector<cv::Point> > contours; cv::findContours(bin_mat.clone(), contours , cv::RETR_EXTERNAL, cv::CHAIN_APPROX_SIMPLE); std::vector<std::vector<cv::Point> > contours2; for(auto itr = contours.begin(); itr != contours.end(); ++itr) { if(cv::contourArea(*itr) > 8) { contours2.push_back(*itr); cv::Rect rc = cv::boundingRect(*itr); cv::rectangle(src_mat, rc, cv::Scalar(0,0,255)); } } cv::Mat ctr_mat = cv::Mat::zeros(bin_mat.size(), CV_8U); cv::drawContours(ctr_mat, contours2, -1, cv::Scalar::all(255), 1); C#:VectorOfVectorOfPoint contours = new VectorOfVectorOfPoint(); VectorOfVectorOfPoint contours2 = new VectorOfVectorOfPoint(); Mat hierarchy; CvInvoke.FindContours(binMat, contours, hierarchy, RetrType.External, ChainApproxMethod.ChainApproxSimple); for (int i = 0; i < contours.Size; ++i) { if (CvInvoke.ContourArea(contours[i]) > 8) { contours2.Push(contours[i]); Rectangle rc = CvInvoke.BoundingRectangle(contours[i]); CvInvoke.Rectangle(srcMat, rc, new MCvScalar(0, 0, 255)); } } Mat ctrMat = new Mat(binMat.Size, DepthType.Cv8U, 3); CvInvoke.DrawContours(ctrMat, contours2, -1, new MCvScalar(255)); The contours have become much smaller, the noise has become worse, but we continue to lose the information and the objects of interest (the automobiles), making them eventually more unlike the original. Those that are close to the camera, nowhere else went, but the more so, the more the situation is getting worse: At the end of the day, it may be necessary to draw the lumbaric contours of rectangles, and then it will be seen what it might be possible to draw together the dispersed parts of each car in a single whole: But the film does not question: parts of different objects are located at relatively same distances as parts of the same object, leading to the impossibility of correct clustering. And since we have other factors that might lead to the binarization, there is simply no problem with the automatic definition of the frontiers of motor vehicles under existing conditions. Well, maybe one or two cars close to the camera, but not anymore.In that case, there is nothing more than manually defining the limits of parking spaces. This task is usually not postponed because it has an arsenic function (with the user pointing out the location of the parking multi-angles) as a whole limited and implemented by means of general-purpose framings.A simple design can be used to obtain a map of the parking space from predetermined coordinates in the frame:cv::Mat parking_lot_mat = src_mat(cv::Rect(0,0,100,100)).clone(); Now. parking_lot_mat will contain part of the reference image at coordinates x equal to 0 and y equal to 0, and will be 100x100 peaks.Then you're doing all the same work on binarization as you did on the big picture. At the end, the contour and its area will be found by means of a method: cv::contourArea()♪ If the amount of the contour area exceeds a threshold shown empirically, then the vehicle in this parking area can be considered detected.

It might help: http://technhard.blogspot.ru/2013/02/web-opencv-linux.html

I'm afraid that's not gonna help with this task by a traffic detector.BackgroundSubtractorMOG2or even more of a cross-sectional difference.The parking is usually the street, which means the weather and the overflow of lighting. The shadows, after all, that make their border with the solar part more contrasting than the color of a car near the grey. And it's still changing in time. The cloud and sun weather with the wind is particularly pronounced. On the basis of the reference, the author of the draft clarifies that different sets of teaching images have been prepared under different weather conditions. Of course, we can't stop mentioning pedestrians, and the drivers themselves, who sniff around cars about and without, and create endless false reactions.Algorithms working with the so-called feature points (specificalities) available in OpenCV are also not working for this task. Even those that are meant to estimate the area. It's complicated.In my view, there's nothing going to help but machine training on pre-prepared samples and a lot of them. It's all on the same Haara that they refused. It's not necessarily this algorithm (so it's likely that it won't work) but any other one who can, on some grounds, catch that there's no such place.SupplementAs an option, we could try to use Cannie's operator. This algorithm is available in the OpenCV toolkit.Take two pictures of the same parking spot with the car and no. Try to pick Kenny's coefficients for both images so that where there is a car, there are substantially more boundaries than that where the car is not (as paradoxically, but at "even, clean, one-coloured" asphalt, under certain conditions, it is possible to detect another messivo from the colour lines.If it doesn't work it out, then you'll get a blur in front of Kenny.Next, if you do, take the other two pictures with the car and the already same parking space, but they're already over time. Like, if the first couple was from the morning shoot, take the evening. For the second pair, try the same coefficients as the first.If you succeed, change the pictures of the parking spot and the car on it. And also try to apply the same coefficients, but now to new images.If there's a patience for everything, it'll take a maximum of a few hours, but it'll be perfectly clear if the path to the colour margin analysis is right or it should be used to use other instruments. This experiment will save time if something goes wrong or approves primary judgements.

The best option is not to guess (and not to make the community know) but to make it a baccalaureate.You've got four or three of mine, one of in the commentaries.Leave it as it is and remove it from the vector. copy list and removed Your option with extra volumeCouple the necessary to the new vector and remove the old one.All options have both pros and cons. Benchmark will.

Probably didn't connect all the required. lib Project files, and they're connected to your separate projects. In the properties of the project, Linker->Input->Additional Dependencies Sign the qt and/or opencv. See what projects are in place.

In brief, the identification of software tools should always be subject to a specific type of facility. If the conversation begins with words, pray, not find, and don't we recognize anything that's just going to get caught in the frame, then she's gonna have a problem. Not invented yet. Even if some software allows for the search for arbitrary objects, it is in fact a variety of decisions and algorithms collected together.Further consideration should be given to the type of object proposed for recognition. What is his geometric form/form. From what angles the film will take. Whether the lighting breaks are strong. Subject in motion or static. Is there a set (usually a kit equivalent to hundreds of thousands and more samples) of images for car training? And many other conditions and nuances. Even a human trivial object like a dog creates many variations. For example, there will be only a face in the frame (anphas, profile, even upside down) or a whole cart full of paint, and what about wool (some kind of polochmatist, some less) and so on.All these differences are fundamentally practical for any recognition algorithm that will be desired. Some of them will be insensitive to the colors and shades, but they will be sensitive to the angle of the film. Some will be insensitive to turns and inclinations, but perfectly sensitive to the shape of the object.Contour analysis is an excellent choice, but, as has already been pointed out on the issue, it is too necessary to have so-called perfect conditions. In the case of a dog, the perfect conditions would be uniformly pouring it all with some paint, like red, and putting it in a room with white floors and walls. Instead of flashing lights, a searchlight shall be installed in the corners of the corners, giving the same and non-flashing light. Afterwards, we need to make an unbearable animal sit peacefully and not move so that the geometric shape of the contours that will be detected in the frame will change marginally. Of course, the software that has been acquired during this experiment will only be able to recognize red dogs sitting in a white room with a viewer. It's a questionable profiling, because it's only a little easier to make them for contour analysis less perfect, like an algorithm would stop working.Contours are a two-way form. This is, in fact, a small number of processing information, which, on the one hand, results in savings in computing resources and an increase in algorithm speeds, but on the other hand, there are no indications without which, in some cases, recognition is simply impossible. Try to guess where the pictures are number one and where 7. Maybe there's two units or two sevens... looking at what angle and what type of print. ♪ ♪It is clear that the contour analysis is not appropriate to recognize such objects as a dog (yes of any one whose shape is not stable) and where objects may be deformed, which is the norm for the natural environment.Well, there's an interesting thing to do: how if the two-worlders tried to look at three-dimensional objects. The cube would be a square for them, and the ball would be round. That's not what the camera sees as being able to simulate the human brain by adding the necessary details that didn't see the eyes on their own. That must be borne in mind.