ENROLL NOW: Computer Vision Basics Course
Week 1: Computer Vision Basic Course Certification Answers: Coursera
Question 1: Computer vision includes which of the following?
- Automatic extraction of features from images
- All are correct
- None are correct
- Understanding useful information
- Analysis of images
Question 2: The image acquisition devices of computer vision systems capture visual information as digital signals?
- True
- False
Question 3: Correct Syntax to read image in MATLAB in current folder
- var_image = imread(‘my_image.jpg’)
- var_image = imread(‘my_image’)
Question 4: Select the correct option to crop top-left 50×50 section from the image read below.
- var_image = imread(‘my_image.jpg’)
- cropped_section = var_image(0:50,0:50)
- cropped_section = var_image(1:50,1:50)
- cropped_section = var_image[0:50,0:50]
- cropped_section = var_image[1:50,1:50]
Question 5: What is initial data type of the image you read through imread function of MATLAB?
- int8
- double
- uint8
Question 6: I1 = imread(‘my_image.jpg’)
I2 = im2double(I1)
- Scales the image intensity values from 0 to 1
- Converts the image from uint8 to double format
- The array dimensions remain same
Question 7: Select the options which assigns height and width of an image correctly in MATLAB.
var_image = imread(‘my_image.jpg’)
- [height,width] = size(var_image );
- image_dimension = size(var_image );
- height = image_dimension(1)
- width = image_dimension(2)
- [width,height] = size(var_image );
- image_dimension = size(var_image );
- width = image_dimension(1)
- height = image_dimension(2)
Question 8: Accessing Image Sub-Regions
img = imread(‘cameraman.tif’);
subimg1 = img(1:50,1:50);
subimg2 = img( end -49 :end, end -49 :end);
SSD = sum(sum((double(subimg1) – double(subimg2)).^2));
SSD = immse(subimg1, subimg2) * numel(subimg1);
disp(SSD);
Week 2: Computer Vision Basic Course Certification Answers : Coursera
Question 9: Which of the following are area sources?
- Bulb
- All of these
- Sun at infinity
- Diffuser boxes
- White walls
Question 10: Does distance of the light source affect the color of a pixel?
- No
- Yes
Question 11: We lose depth information in perspective projection.
- True
- False
Question 12: Match column A with correct options in column in B
| Column A | Column B |
| 1) Shutter speed | a) Amount of light per unit area reaching image sensor |
| 2) Exposure | b) an effect that causes different signals to become indistinguishable when sampled |
| 3) Aperture | c) The length of time when sensor is exposed to light when taking a photograph |
| 4) Aliasing | d) Hole or an opening through which light travels |
Answer: 1-c, 2-a, 3-d, 4-b
Question 13: Color Imaging – RGB Channels
%Read the image
img = imread(‘image.jpg’);
%Get the size (rows and columns) of the image
[r,c] = size(img);
rr=r/3;
%Wrire code to split the image into three equal parts and store them in B, G, R channels
B=imcrop(img,[1,1,c,rr-1]);
G=imcrop(img,[1,1*rr+1,c,rr-1]);
R=imcrop(img,[1,2*rr+1,c,rr]);
%concatenate R,G,B channels and assign the RGB image to ColorImg variable
ColorImg(:,:,1) = R;
ColorImg(:,:,2) = G;
ColorImg(:,:,3) = B;
imshow(ColorImg)
Week 3: Computer Vision Basic Course Certification Answers : Coursera
Three-Level Paradigm
Question 14:
| Column A | Column B |
| 1) Computational Theory | a) Steps for Computation |
| 2) Representation and algorithm | b) Physical realization of algorithms, programs and hardware |
| 3) Implementation | c) What the device is supposed to do |
Answer: 1-c, 2-a, 3-b
Question 15: Low-level vision consists of:
1) feature detection and matching
2) early segmentation
- 1
- 1 and 2
- 2
- None
Question 16: Image Gradient Magnitude
img = imread(‘cameraman.tif’);
[Gx, Gy] = imgradientxy(img); [Gmag, Gdir] = imgradient(Gx, Gy);%Uncomment the code below to visualize Gx and Gy
%imshowpair(Gx,Gy,’montage’)
%Uncomment the code below to visualize Gmag and Gdir
%imshowpair(Gmag,Gdir,’montage’)
Week 4: Computer Vision Basic Course Certification Answers : Coursera
Question 17: Match the Algorithms in column A with correct techniques in column B
| Column A | Column B |
| 1) Dynamic Programming | a) Binary Image Restoration |
| 2) Graph algorithms | b) Stereo matching |
| 3) Dynamic Programming | c) Seam Carving |
| 4) Graph algorithms | d) Image segmentation |
Answers: 1-b, 2-d, 3-c, 4-a
Question 18: Aligning RGB Channels (using SSD)
img = imread(‘course1image.jpg’);
[height, width] = size(img);
oneThird = floor(height/3);
B = img(1:(oneThird), :);
G = img((oneThird+1):(2*oneThird), :);
R = img((2*oneThird+1):(3*oneThird), :);
c_x = (341/2-25);
c_y = (400/2-25);
ref_img_region = double(G(c_x:c_x + 50, c_y:c_y + 50));
red_offset = offset(double(R(c_x:c_x + 50, c_y:c_y + 50)), ref_img_region);
shifted_red = circshift(R, red_offset);
blue_offset = offset(double(B(c_x:c_x + 50, c_y:c_y + 50)), ref_img_region);
shifted_blue = circshift(B, blue_offset);
ColorImg_aligned = cat(3, shifted_red, G, shifted_blue);
%ColorImg_aligned = cat(3, G, shifted_red, shifted_blue);
%ColorImg_aligned = cat(3, G, shifted_blue, shifted_red);
%ColorImg_aligned = cat(3, G, shifted_blue, shifted_red);
imshow(ColorImg_aligned);
% Find the minimun offset by ssd
function [output] = offset(img1, img2)
MIN = inf;
for x = -10:10
for y = -10:10
temp = circshift(img1, [x, y]);
ssd = sum((img2 – temp).^2, ‘all’);
if ssd < MIN
MIN = ssd;
output = [x, y];
end
end
end
end
Priya Dogra – Certification | Jobs | Internships