An input image has the size of 140x140 and we use a filter (kernel) with the size of 20x20. We then convolve the input image with the filter. a. If we do not use padding and stride is 1, the output size will be axa, where a is
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Q: An input image has the size of 140x140 and we use a filter (kernel) with the size of 20x20. We then…
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Q: An input image has the size of 140x140 and we use a filter (kernel) with the size of 20x20. We then…
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A: Answer: we have explain in very well as my expected in brief explained
Q: An input image has the size of 140x140 and we use a filter (kernel) with the size of 20x20. We then…
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Q: An input image has the size of 140x140 and we use a filter (kernel) with the size of 20x20. We then…
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Q: An input image has the size of 140x140 and we use a filter (kernel) with the size of 20x20. We then…
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- For an image of size 32x32 and a filter of size 5x5, if you want the convolution result to keep the original image size, what's the padding size in pixels? a) 5 b) 2 c)10 d) 3 explain.The input image has been converted into a matrix of size 28 X 28 and a kernel/filter of size 3 X 3 with a stride of 2 and padding of 1. What will be the size of the convoluted matrix? A. 15 x 15 B. 14 x 14 C. 14 x 15 D. 14.5 x 14.5"""You are given an n x n 2D mat representing an image. Rotate the image by 90 degrees (clockwise). Follow up:Could you do this in-place?""" # clockwise rotate# first reverse up to down, then swap the symmetry# 1 2 3 7 8 9 7 4 1# 4 5 6 => 4 5 6 => 8 5 2# 7 8 9 1 2 3 9 6 3 def rotate(mat): if not mat: return mat mat.reverse() for i in range(len(mat)): for j in range(i): mat[i][j], mat[j][i] = mat[j][i], mat[i][j] return mat if __name__ == "__main__": mat = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] print(mat) rotate(mat) print(mat).
- Computer Science For an image represented by a given large 3d numpy array, use numpy to get a neighborhood filter using the mean of the pixels. For example, look through each of the image’s 3×3 window of pixels and replace outlier pixels with the mean value of the others in the 3×3 window. Set the resulting pixel value to be the mean value of a pixel and its eight neighbors. For pixels on an edge, use the mean value of the pixel and its five neighbors. For pixels on corners, use the mean value of the pixel and its three neighbors. If you loop over every pixel then you must store your input and output arrays separatelyIt will simulate the diffusion of heatthrough a square metal plate. The plate starts off at 0 deg., but the 4 ×4 pixel square in the centeris kept at 100 deg.Now a cell in the interior has 4 neighbors: the upper, lower, left, and right neighbor. The updateformula for an interior point (other than the center square) is therefore given by:u′i,j = ui,j + k((ui,j+1 −ui,j ) + (ui,j−1 −ui,j ) + (ui−1,j −ui,j ) + (ui+1,j −ui,j ))= ui,j + k(ui,j+1 + ui,j−1 + ui−1,j + ui+1,j −4ui,j )The boundaries have to be updated separately using 8 different equations: 4 for each of the edges,and 4 for each corner. Here is the update formula for a cell on the top edge that is not on a corner:u′i,n−1 = ui,n−1 + k((ui,n−2 −ui,n−1) + (ui−1,n−1 −ui,n−1) + (ui+1,n−1 −ui,n−1))= ui,n−1 + k(ui,n−2 + ui−1,n−1 + ui+1,n−1 −3ui,n−1)Here is the update formula for a cell in the top left corner:u′0,n−1 = u0,n−1 + k((u0,n−2 −u0,n−1) + (u1,n−1 −u0,n−1))= u0,n−1 + k(u0,n−2 + u1,n−1 −2u0,n−1) Program:For an object detection problem, assume you are designing a YOLO like model to do the job. Your input image size is 127x127 (RGB). We are looking for a 3x3 output grid size. The number of classes is 20 and for each cell in the grid, we are considering 2 anchors. Design the CNN network and as a designer feel free to set your network's hyper-parameters as you wish. The output of your network needs to show the final height, width, and the channel. You do not need to include post-processing (nan-max suppression) step for YOLO. In Python-
- While doing Histogram Stretching, we established that we go pixel by pixel through whole image and apply Histogram Stretching Formula for each pixel. So, for example in the following image, if there are 100 pixels, we would do Histogram Stretching Formula 100 times. Do you think we could do Histogram Stretching for whole image and calculate this formula fewer times ? If no why ? If yes, how ?In attached image, there are 5 states, a, b, c, d, e. Two actions are available for each state: East, West except for the exit states a and e, where the only action available is “Exit”. The transition is deterministic. The rewards of the exit states are given as shown in Image. A) For γ= 1, what is the true utility ? (Please fill the form completely) Example response format: 10 10 10 10 10 (Please note the space!) B) For γ = 0.1, what is the true utility? Example response format: 10 0.1 10 10 0.1 (Please pay attention to the space!) C) For which γ are West and East equally good at state d? (please take to the fourth decimal place)Example response format: γ = 0.1234 (take to the fourth decimal place, please pay attention to the space!)Fix a marker on a wall or a flat vertical surface. From a distance D, keeping the camera stationed static (not handheld and mounted on a tripod or placed on a flat surface), capture an image such that the marker is registered. Then translate the camera by T units along the axis parallel to the ground (horizontal) and then capture another image, with the marker being registered. Compute D using disparity based depth estimation in stereo-vision theory
- subject : smart computing To animate an object while making the movements look realistic, we need to take care of acceleration and deceleration. With nonlinear interpolation, we can do that by using varying time steps using constant time steps fixing the number of inter between frames reducing the number of inter between framesThe question is on the attached image. grayscale function code: from images import Image def grayscale1(image): """Converts an image to grayscale using the psychologically accurate transformations.""" for y inrange(image.getHeight()): for x inrange(image.getWidth()): (r, g, b) = image.getPixel(x, y) r = int(r * 0.299) g = int(g * 0.587) b = int(b * 0.114) lum = r + g + b image.setPixel(x, y, (lum, lum, lum))Describe an algorithm for converting a greyscale image to a two-bit image (i.e. only using black, dark grey, light grey and white), while preserving as much visual quality as possible. Your algorithm should map every input greyscale pixel to exactly one output pixel.