EE 168 Introduction to Digital Image Processing January 30, 2012 HOMEWORK 2 SOLUTIONS

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1 EE 168 Introduction to Digital Image Processing January 3, 212 HOMEWORK 2 SOLUTIONS Problem 1: Image Scaling We determine the dimensions of the original image (read from the rodin.raw file) to be 348x261. Figure 1-1 shows a zoomed-in-portion of this image after it is zero-filled-scaled by factors of 2, 3, and 4. Note that as the scale factor increases, more of the pixels are assigned to a value of zero due to zero-filling. As a result, scaling an image by a large factor results in low quality images. Several techniques, including zero-order and first-order hold, can be used to improve image quality. In the zero-order hold algorithm, each zero-filled pixel is replaced by the closest preceding non-zero pixel value. The resulting images are shown in Figure 1-2. Since each zerofilled pixel takes the value of the closest non-zero preceding pixel, repetition of each pixel value for n times (where n is the scaling factor) is observed in cuts through features in the image. Figure 1-3 demonstrates this for the feature located at line and columns 6-7 in the original image. The first-order hold is different in that each zero-filled pixel takes the value of a weighted average of the closest two pixels. The weights are determined by the distance between the zerofilled pixel and the non-zero pixels. Due to this averaging operation, the cuts through features show a piecewise linear variation (Figure 1-5), rather than the piecewise constant variation that we saw with the zero-order hold in Figure 1-3. Note that each of these linear pieces ext for n values (where n is the scaling factor). Original Image Scaled (Zero-fill) by Scaled (Zero-fill) by Scaled (Zero-fill) by Figure 1-1: Effect of Zero-filling on Image Quality: zoomed portions of the scaled images 1

2 Original Image Scaled (Zero-order hold) by Scaled (Zero-order hold) by 3 Scaled (Zero-order hold) by Figure 1-2: Zero-Order hold images 2 Original image 2 Scaled (Zero-order hold) by Scaled (Zero-order hold) by 3 2 Scaled (Zero-order hold) by Figure 1-3: Cuts through a feature (at line, columns 6-7 in the original image) showing change of pixel values 2

3 Original Image Scaled (First-order hold) by Scaled (First-order hold) by 3 Scaled (First-order hold) by Figure 1-4: First-Order hold images 2 Original image 2 Scaled (First-order hold) by Scaled (First-order hold) by 3 2 Scaled (First-order hold) by Figure 1-5: Cuts through a feature (at line, columns 6-7 in the original image) showing change of pixel values 3

4 Problem 2: Image Subsetting and Translation Figure 2-1 shows a subset image of dimensions 1 by 18 from the original rodin.raw image and its insertion into three different locations of a 3 by 4 image. Note that the insertion replaces the corresponding pixel values of the 3 by 4 image with the pixel values of the inserted subset image. Subset Image Figure 2-1: Image Subsetting and Translation Problem 3: Rotations The original image and its rotated versions for rotation angles of 45,, and 67 degrees are shown in Figure 3-1. Note that the rotation is performed by changing the location of the pixel values using the 2 by 2 rotation transformation matrix, provided in the lecture notes. Note that if the origin of the image coordinates is in the upper left corner (and the coordinates are thus increasing downward), the transformation matrix derived in class will operate on the [rowindex column-index] vector rather than the [x y] vector. The rotation problem is different form the other problems in that we are changing the location of the pixels rather than the values that the pixels take. It is observed that the rotation causes black spots in the images due to the rounding that takes place in the algorithm. These spots can be removed by applying techniques similar to a zero-order or first order hold. As an example, the black spots are replaced by the value of the closest preceding non-zero pixel value, as shown in Figure 3-2. Note that the visual quality is significantly higher in Figure 3-2 than it is in Figure

5 Original Image 7 45 Degree Rotation Degree Rotation 67 Degree Rotation Figure 3-1: Image Rotation Original Image 7 45 Degree Rotation Degree Rotation 67 Degree Rotation Figure 3-2: Image Rotation with Zero-Order Hold 5

6 Problem 4: Adding Images Together A subset (dimensions 1 by 18) is taken from the original image and is pasted at three different locations in the original image, as shown in Figure 4-1. Figure 4-2 shows the results from bling the subset image with the original image at the three different locations. Bling is done by assigning the average of the original pixel value and the subset pixel value. Subset Image Figure 4-1: Image Subsetting Subset Image is pasted at 3 different locations within the original Subset Image Figure 4-2: Image Subsetting and Bling Subset Image is bled with the original 6

7 Problem 5: Subtracting to get image change The two original images (from lab2prob5data1 and lab2prob5data2 ) are shown in Figure 5-1. The difference of the two images is shown in Figure 5-2. We see that the two images are mostly similar, except for possibly a slight translation or difference in general illumination levels. The primary difference between the two images is the location of the two cars. These differences are captured and highlighted in the difference image. Hence, difference images can be particularly helpful in finding differences between two images that appear visually alike. Original Image-1 Original Image Figure 5-1: Original Images Difference Image Figure 5-2: Difference Image 7

8 Problem 6: Image Inversion Figure 6-1 compares the image (from lab2prob6data ) and its inverse. Note that the inverse of the image is obtained by subtracting it from 255 since we are using 256 possible gray levels, ranging from to 255. Original Image Inverse Image Figure 6-1: Image Inversion Problem 7: An Unknown Format After opening the file lab2prob7data, we find that it has elements. We know that there is a main header at the beginning of the file, header information at the beginning of each line, and trailer information at the of the file. Because the line-header information will repeat at the beginning of each line, we can find the line-length by plotting the first few thousand elements of the raw data and looking for a repetitive pattern, as shown in Figure 7-1. We note that the line-header repeats every 582 elements, thus the line-length is 582. By zooming in on the beginning of each new cycle, we observe that the line header is 36 elements long and is filled with zeros, except for the 24 th element, which contains line count information. This implies an actual image data length of 546 (582 36). Working backwards from the start of the second line (at element 727), we find that the first line of the image starts at element 145 ( ). This means that we can skip over the first 144 elements (the beginning main header) of the file to get to the image + line-header data. Note that it seems some of the beginning main-header information overlaps with the first line s header. From the line count information contained in column 24 of each line header, we find that there are 691 image lines. Furthermore, if we zoom in on the of the file, we see that the trailer consists of an extra line of zeros. We can then trim out the 36 line-header columns and the 692 nd row to get the final image of Yoda from the Star Wars films, shown in Figure 7-2, which has dimensions 691x546. The final image, with all extraneous bytes removed from the data stream, is then converted to a common raster format (such as tiff, jpeg, gif, etc.). 8

9 2 First elements of raw data Figure 7-1: Plot of first elements in file; reveals the line-length and line-header length Figure 7-2: Final Image of Yoda, with extraneous data removed 9

10 MATLAB code: Question 1: Image Scaling fid = fopen('rodin.raw'); im = fread(fid, [ ], 'uint8'); im = im'; fclose(fid); numrow = size(im, 1); numcol = size(im, 2); %Zero-filled: scale by 2 im_zerofill2 = zeros(2*numrow, 2*numcol); im_zerofill2(1:2:2*numrow, 1:2:2*numcol) = im; %Zero-filled: scale by 3 im_zerofill3 = zeros(3*numrow, 3*numcol); im_zerofill3(1:3:3*numrow, 1:3:3*numcol) = im; %Zero-filled: scale by 4 im_zerofill4 = zeros(4*numrow, 4*numcol); im_zerofill4(1:4:4*numrow, 1:4:4*numcol) = im; % Zero-Order Hold % In the zero-order hold for a scaling factor of n the following steps are taken: % i) form a new all zeros matrix that is nxn times larger than the original matrix. % ii) set every (nxm)th row, (nxl) column element of the new matrix to the mth row, lth column % element of the original matrix % iii) Copy the value in the (nxm)th row, (nxl)th column of this new matrix to its elements in % the (nxm):(nx(m+1))-1 rows and (nxl):(nx(l+1))-1 columns %Zero-order hold: scale by 2 im_zerohold2 = zeros(2*numrow, 2*numcol); im_zerohold2(1:2:2*numrow, 1:2:2*numcol) = im; im_zerohold2(2:2:2*numrow,:) = im_zerohold2(1:2:2*numrow,:); im_zerohold2(:, 2:2:2*numcol) = im_zerohold2(:, 1:2:2*numcol); %Zero-order hold: scale by 3 im_zerohold3 = zeros(3*numrow, 3*numcol); im_zerohold3(1:3:3*numrow, 1:3:3*numcol) = im; im_zerohold3(2:3:3*numrow,:) = im_zerohold3(1:3:3*numrow,:); im_zerohold3(3:3:3*numrow,:) = im_zerohold3(1:3:3*numrow,:); im_zerohold3(:, 2:3:3*numcol) = im_zerohold3(:, 1:3:3*numcol); im_zerohold3(:, 3:3:3*numcol) = im_zerohold3(:, 1:3:3*numcol); %Zero-order hold: scale by 4 im_zerohold4 = zeros(4*numrow, 4*numcol); im_zerohold4(1:4:4*numrow, 1:4:4*numcol) = im; im_zerohold4(2:4:4*numrow,:) = im_zerohold4(1:4:4*numrow,:); im_zerohold4(3:4:4*numrow,:) = im_zerohold4(1:4:4*numrow,:); im_zerohold4(4:4:4*numrow,:) = im_zerohold4(1:4:4*numrow,:); im_zerohold4(:, 2:4:4*numcol) = im_zerohold4(:, 1:4:4*numcol); im_zerohold4(:, 3:4:4*numcol) = im_zerohold4(:, 1:4:4*numcol); im_zerohold4(:, 4:4:4*numcol) = im_zerohold4(:, 1:4:4*numcol); % First-Order Hold % In the first-order hold for a scaling factor of n the following steps are taken: % i) form a new all zeros matrix that is nxn times larger than the original matrix. 1

11 % ii) set every (nxm)th row, (nxl) column element of the new matrix to the mth row, lth % column element of the original matrix % % iii) Zero-filled pixels are replaced by a weighted average of the closest % neighboring pixels. % For a scaling of 2, [a b] becomes [a.5*a+.5*b b] % For a scaling of 3, [a b] becomes [a (2/3)*a+(1/3)*b (1/3)*a+(2/3)*b b] % For a scaling of 4, [a b] becomes [a (3/4)*a+(1/4)*b (2/4)*a+(2/4)*b (1/4)*a+(3/4)*b b] %First-order hold: scale by 2 im_firsthold2 = zeros(2*numrow, 2*numcol); im_firsthold2(1:2:2*numrow, 1:2:2*numcol) = im; im_firsthold2(2:2:2*numrow,:) =.5*im_firsthold2(1:2:2*numrow,:) +.5*[im_firsthold2(3:2:2*numrow,:); zeros(1, 2*numcol)]; im_firsthold2(:, 2:2:2*numcol) =.5*im_firsthold2(:, 1:2:2*numcol) +.5*[im_firsthold2(:, 3:2:2*numcol) zeros(2*numrow, 1)]; %First-order hold: scale by 3 im_firsthold3 = zeros(3*numrow, 3*numcol); im_firsthold3(1:3:3*numrow, 1:3:3*numcol) = im; im_firsthold3(2:3:3*numrow,:) = (2/3)*im_firsthold3(1:3:3*numrow,:) + (1/3)*[im_firsthold3(4:3:3*numrow,:); zeros(1, 3*numcol)]; im_firsthold3(3:3:3*numrow,:) = (1/3)*im_firsthold3(1:3:3*numrow,:) + (2/3)*[im_firsthold3(4:3:3*numrow,:); zeros(1, 3*numcol)]; im_firsthold3(:, 2:3:3*numcol) = (2/3)*im_firsthold3(:, 1:3:3*numcol) + (1/3)*[im_firsthold3(:, 4:3:3*numcol) zeros(3*numrow, 1)]; im_firsthold3(:, 3:3:3*numcol) = (1/3)*im_firsthold3(:, 1:3:3*numcol) + (2/3)*[im_firsthold3(:, 4:3:3*numcol) zeros(3*numrow, 1)]; %First-order hold: scale by 4 im_firsthold4 = zeros(4*numrow, 4*numcol); im_firsthold4(1:4:4*numrow, 1:4:4*numcol) = im; im_firsthold4(2:4:4*numrow,:) = (3/4)*im_firsthold4(1:4:4*numrow,:) + (1/4)*[im_firsthold4(5:4:4*numrow,:); zeros(1, 4*numcol)]; im_firsthold4(3:4:4*numrow,:) = (2/4)*im_firsthold4(1:4:4*numrow,:) + (2/4)*[im_firsthold4(5:4:4*numrow,:); zeros(1, 4*numcol)]; im_firsthold4(4:4:4*numrow,:) = (1/4)*im_firsthold4(1:4:4*numrow,:) + (3/4)*[im_firsthold4(5:4:4*numrow,:); zeros(1, 4*numcol)]; im_firsthold4(:, 2:4:4*numcol) = (3/4)*im_firsthold4(:, 1:4:4*numcol) + (1/4)*[im_firsthold4(:, 5:4:4*numcol) zeros(4*numrow, 1)]; im_firsthold4(:, 3:4:4*numcol) = (2/4)*im_firsthold4(:, 1:4:4*numcol) + (2/4)*[im_firsthold4(:, 5:4:4*numcol) zeros(4*numrow, 1)]; im_firsthold4(:, 4:4:4*numcol) = (1/4)*im_firsthold4(:, 1:4:4*numcol) + (3/4)*[im_firsthold4(:, 5:4:4*numcol) zeros(4*numrow, 1)]; %Plot the zero-fill results figure; colormap(gray); subplot(2,2,1); imagesc(im); axis([ 7 ]); axis('equal') title('original Image') subplot(2,2,2); imagesc(im_zerofill2); axis([ 7 ]); axis('equal'); title('scaled (Zero-fill) by 2') subplot(2,2,3); imagesc(im_zerofill3); axis([ 7 ]); axis('equal'); title('scaled (Zero-fill) by 3') subplot(2,2,4); imagesc(im_zerofill4); axis([ 7 ]); axis('equal'); 11

12 title('scaled (Zero-fill) by 4') %Plot the Zero-order hold results figure; colormap(gray) subplot(2,2,1); imagesc(im); axis('equal') title('original Image') subplot(2,2,2); imagesc(im_zerohold2); axis('equal'); title('scaled (Zero-order hold) by 2') subplot(2,2,3); imagesc(im_zerohold3); axis('equal'); title('scaled (Zero-order hold) by 3') subplot(2,2,4); imagesc(im_zerohold4); axis('equal'); title('scaled (Zero-order hold) by 4') %Plot Cuts through the zero-order-hold scaled images cut_row = ; cut_col = 6; cut_width = 1; figure; colormap(gray) subplot(2,2,1); plot(im(cut_row, cut_col:cut_col+cut_width)); title('original image') subplot(2,2,2); plot(im_zerohold2(2*cut_row-1, (2*cut_col-1):(2*cut_col-1)+2*cut_width)); title('scaled (Zero-order hold) by 2') subplot(2,2,3); plot(im_zerohold3(3*cut_row-2, (3*cut_col-2):(3*cut_col-2)+3*cut_width)); title('scaled (Zero-order hold) by 3') subplot(2,2,4); plot(im_zerohold4(4*cut_row-3, (4*cut_col-3):(4*cut_col-3)+4*cut_width)); title('scaled (Zero-order hold) by 4') %Plot the First-order hold figure; colormap(gray) subplot(2,2,1); imagesc(im); axis('equal') title('original Image') subplot(2,2,2); imagesc(im_firsthold2); axis('equal'); title('scaled (First-order hold) by 2') subplot(2,2,3); imagesc(im_firsthold3); axis('equal'); title('scaled (First-order hold) by 3') subplot(2,2,4); imagesc(im_firsthold4); axis('equal'); title('scaled (First-order hold) by 4') %Plot Cuts through the first-order-hold scaled images cut_row = ; cut_col = 6; cut_width = 1; figure; colormap(gray) subplot(2,2,1); plot(im(cut_row, cut_col:cut_col+cut_width)); title('original image') subplot(2,2,2); plot(im_firsthold2(2*cut_row-1, (2*cut_col-1):(2*cut_col-1)+2*cut_width)); title('scaled (First-order hold) by 2') subplot(2,2,3); plot(im_firsthold3(3*cut_row-2, (3*cut_col-2):(3*cut_col-2)+3*cut_width)); title('scaled (First-order hold) by 3') subplot(2,2,4); plot(im_firsthold4(4*cut_row-3, (4*cut_col-3):(4*cut_col-3)+4*cut_width)); title('scaled (First-order hold) by 4') 12

13 Question 2: Image Subsetting and Translation fid=fopen('rodin.raw'); im = fread(fid,[ ],'uint8'); fclose(fid); im=im'; im_subset=im(51:,11:28); figure; subplot(2,2,1); colormap('gray'); imagesc(im_subset); title('subset Image') axis('image') f=fopen('lab2prob5data1'); im_b =fread(f,[64 37/64],'uint8'); im_b=im_b'; subplot(2,2,2) im_c = im_b(1:3,1:4); im_c(1:1,1:18)=im_subset; imagesc(im_c) subplot(2,2,3) im_c = im_b(1:3,1:4); im_c(11:2,11:28)=im_subset; imagesc(im_c) subplot(2,2,4) im_c = im_b(1:3,1:4); im_c(151:3,11:28)=im_subset; imagesc(im_c) Question 3: Rotations % rot_matrix is the 2 by 2 rotation matrix discussed in class % Pixel locations are changed using rot_matrrix applied to [rot_ind and col_ind]; % round command and an offset of twice the maximum image dimension is used to % avoid fractional or negative indices. % Rotation for 45 degrees is done below. Rotation for the other angles % can be obtained by changing the rot_angle value clear; n_col = 348; n_row = 261; %dimensions of rodin.raw image fid=fopen('rodin.raw'); im = fread(fid,[n_col n_row],'uint8'); fclose(fid); im = im'; figure; imagesc(im); colormap(gray); rot_angle = 45*pi/18; %in radians %rot_angle = *pi/18; %rot_angle = 67*pi/18; rot_mat = [cos(rot_angle) sin(rot_angle); -sin(rot_angle) cos(rot_angle)]; %initialize to negative 1's to distinguish from the black spots in the image offset = 2*max(size(im)); im_rot=-ones(4*max(size(im)), 4*max(size(im))); for row_ind=1:n_row, for col_ind=1:n_col, im_rot(round(rot_mat(1,1)*row_ind+rot_mat(1,2)*col_ind + offset), 13

14 round(rot_mat(2,1)*row_ind+rot_mat(2,2)*col_ind) + offset)=im(row_ind,col_ind); figure; imagesc(im_rot); colormap(gray); axis('equal'); axis('image'); title(['image rotated by ', num2str(rot_angle*18/pi), ' degrees']) % The following part is the zero-order hold operation to avoid % the black spots in the rotated image. im_rot_filled = im_rot; %initialize to the holey rotated image flag_imagerow = zeros(1, size(im_rot,1)); %keep track of which rows the image is on flag_lastimagecol = zeros(1, size(im_rot, 1)); %keep track of last image column for each row for row_ind = 1:size(im_rot, 1), for col_ind = 1:size(im_rot, 2), if im_rot(row_ind, col_ind) > -1, flag_imagerow(row_ind) = 1; flag_lastimagecol(row_ind) = col_ind; %will overwrite until reaches last col %fill the image with the nearest 'non-zero' pixel value to the left along the row for row_ind = 1:size(im_rot, 1), if flag_imagerow(row_ind) == 1, for col_ind = 2:size(im_rot, 2), if im_rot_filled(row_ind, col_ind) == -1, if col_ind < flag_lastimagecol(row_ind), im_rot_filled(row_ind, col_ind) = im_rot_filled(row_ind, col_ind-1); figure; imagesc(im_rot_filled); colormap(gray); axis('equal'); axis('image') title(['image rotated by ', num2str(rot_angle*18/pi), ' degrees, with zero-order hold']) Question 3 Alternative Rotation Methods: % Four Methods are given for possible Matlab image rotation operations: % - two consider improving efficiency by working with matrices % - the other two use for-loops % - two consider forward-mapping from original image coordinate space to rotated image % coordinate space (but then require zero-order hold operations to fill the hole) % - the other two consider inverse mapping to avoid holes in the first place % rot_matrix is the 2 by 2 rotation matrix discussed in class % Pixel locations are changed using rot_matrix applied to [rot_ind and col_ind]; % Alternatively, the inverse of rot_matrix can be applied to the rotated image coordinate space % to find the mapping back to the original coordinate space this method will prevent holes % from developing in the rotated image % Round command and an offset of 3 is used to avoid fractional or negative indices. % Rotation for 45 degrees is done below. Rotation for the other angles % can be obtained by changing the rot_angle value rot_angle = 45*pi/18; %in radians rot_mat = [cos(rot_angle) sin(rot_angle); -sin(rot_angle) cos(rot_angle)]; rot_mat_inv = inv(rot_mat); 14

15 % Method 1: Use matrix operations to map original image coordinates % to rotated image coordinates (will give holes) x = repmat(1,1,n_col); for ind=2:n_row; x = horzcat(x, repmat(ind,1,n_col)); y = repmat([1:n_col],1,n_row); x_y = [x; y]; x_y_prime = rot_mat*x_y; x_y_prime = round(x_y_prime) + 3; %initialize to negative 1's to distinguish from the black spots in the image im_rot = -ones(3*n_col, 3*n_row); for ind1 = 1:length(x_y_prime), im_rot(x_y_prime(1,ind1), x_y_prime(2,ind1)) = im(x_y(1,ind1), x_y(2,ind1)); % Method 2: Use matrix operations to map rotated image coordinates % to the orignal image coordinates (inverse mapping that will prevent holes) im_rot=-ones(3*n_col, 3*n_row); x_pr = repmat(1,1,size(im_rot,2)); for ind=2:size(im_rot,1); x_pr = horzcat(x_pr, repmat(ind,1,size(im_rot,2))); y_pr = repmat([1:size(im_rot,2)],1,size(im_rot,1)); x_y_pr = [x_pr-3; y_pr-3]; x_y = rot_mat_inv*x_y_pr; for ind1 = 1:length(x_y), x = round(x_y(1,ind1)); y = round(x_y(2,ind1)); if(x > & y > & x <= size(im,1) & y <= size(im,2)), im_rot(x_y_pr(1,ind1)+3, x_y_pr(2,ind1)+3) = im(x, y); % Method 3: Use For loop and forward mapping of coordinates % initialize to negative 1's to distinguish from the black spots in the image im_rot=-ones(3*348,3*261); for i=1:n_row, for j=1:n_col, im_rot(round(rot_mat(1,1)*i+rot_mat(1,2)*j)+3,round(rot_mat(2,1)*i+rot_mat(2,2)*j)+3) = im(i,j); %holes should then be filled with a zero-order hold operation % Method 4: Use for-loop and inverse mapping of coordinates (avoids holes) im_rot=-ones(3*n_col, 3*n_row; for i=1:size(im_rot,1), for j=1:size(im_rot,2), %subtract 3 to account for possible negative indices i_prime = round(rot_mat_inv(1,1)*(i-3)+rot_mat_inv(1,2)*(j-3)); j_prime = round(rot_mat_inv(2,1)*(i-3)+rot_mat_inv(2,2)*(j-3)); if (i_prime > & j_prime > & i_prime <= size(im,1) & j_prime <= size(im,2)), im_rot(i,j)=im(i_prime,j_prime); 15

16 Question 4: Adding Images together fid=fopen('rodin.raw'); im = fread(fid,[ ],'uint8'); fclose(fid); im = im'; %take a subset of the original image im_subset = im(51:,11:28); %create three new images with the subset pasted in at different locations im_2 = im; im_2(1:1,1:18)=im_subset; im_3 = im; im_3(11:2,11:28)=im_subset; im_4 = im; im_4(112:261,169:348)=im_subset; figure; colormap('gray'); subplot(2,2,1); imagesc(im_subset); title('subset Image'); axis('image'); subplot(2,2,2); imagesc(im_2); subplot(2,2,3); imagesc(im_3) subplot(2,2,4); imagesc(im_4) % the following part bls the pasted image with the original one. im_2_bl = im; im_2_bl(1:1,1:18)=.5*(im_2_bl(1:1,1:18)+im_subset); im_3_bl = im; im_3_bl(11:2,11:28)=.5*(im_3_bl(11:2,11:28)+im_subset); im_4_bl = im; im_4_bl(112:261,169:348)=.5*(im_4_bl(112:261,169:348)+im_subset); figure; colormap('gray'); subplot(2,2,1); imagesc(im_subset); title('subset Image'); axis('image'); subplot(2,2,2); imagesc(im_2_bl); subplot(2,2,3); imagesc(im_3_bl); subplot(2,2,4); imagesc(im_4_bl); 16

17 Question 5: Subtracting to get image change %note that the sum of the values in the first line of the file gives the line-length fid=fopen('lab2prob5data1'); im_1 = fread(fid,[64 37/64],'uint8'); fclose(fid); im_1 = im_1'; fid=fopen('lab2prob5data2'); im_2 = fread(fid,[64 37/64],'uint8'); fclose(fid); im_2 = im_2'; figure; colormap(gray); subplot(1,2,1); imagesc(im_1); axis('image'); title('original Image-1') subplot(1,2,2); imagesc(im_2) axis('image'); title('original Image-2') figure; colormap('gray'); imagesc(im_1 - im_2); %Difference Image axis('image'); title('difference Image'); Question 6: Image Inversion % The image is inverted by subtracting all of its pixel values from 255. fid=fopen('lab2prob6data'); im=fread(fid,[48 37/48],'uint8'); fclose(fid); im = im'; figure; colormap('gray'); subplot(1,2,1); imagesc(im); axis('equal'); axis('image'); title('original Image') subplot(1,2,2) imagesc(255-im) axis('image'); title('inverse Image') Question 7: An Unknown Format clear fid=fopen('lab2prob7data'); data = fread(fid,[1 inf],'uint8'); fclose(fid); %data is size 1x42741 % Find the line length by looking for a repetitive pattern in the data % The line-header information will repeat at the beginning of each new line figure; plot(data(1:)); title('first elements of raw data') 17

18 line_length = 582; len_line_header = 36; len_main_header = 144; fid=fopen('lab2prob7data'); main_header = fread(fid,[1 len_main_header],'uint8'); %beginning main header is length start reading image after im_raw = fread(fid,[line_length inf], 'uint8'); %raw image has dimensions 692x582 fclose(fid); %find the number of data lines: subtract 1 since the last row is the special trailer, %which we know from observing the line count information contained in col 24 of the line-headers num_lines = size(im_raw,2) - 1; %form image without extraneous data by cutting out line-header columns and trailer %the final image has dimensions of 691x546 im = im_raw(len_line_header+1:, 1:num_lines)'; figure; imagesc(im); colormap(gray) axis('equal') axis('image') % im is 8-bit grayscale (-255) defined as double precision % imwrite() assumes doubles to be in the dynamic range [,1], so convert it to uint8 im_final = uint8(im); imwrite(im_final, 'Yoda.jpeg'); 18