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GEORGIA INSTITUTE OF TECHNOLOGY
School of Electrical and Computer Engineering
ECE 6258 Digital Image Processing Fall 2003
Problem Set #1 – Solutions
Problem 1.1:
The following solution is from Sarju Vaz.
(a)
clear all, close all
% open the image [I, cmap] = imread('cameraman.tiff'); figure, subplot(221), imshow(I) I = double(I);
min_I = min(I(:)) max_I = max(I(:)) orig_num_levels = max_I - min_I + 1
% Quantize the image to 10 uniformly spaced gray levels quant_num_levels = 10
%size of quantization bin (cover for even bin size) quant_bin_size = orig_num_levels/quant_num_levels
xlabel(sprintf(... 'pixel values range from %d to %d\nNum gray levels in orig image = %d\nNum quantization levels = %d\nQuantization bin width ~ %.2f',... min_I, max_I, orig_num_levels, quant_num_levels, quant_bin_size))
%Go with ceil(quant_bin_size) as long as [quant_bin_size*ceil(quant_bin_size)] <= 256
subplot(222), imhist(uint8(I))
bin_size = ceil(quant_bin_size) mid_val = floor(bin_size/2) % more intuitive --> ceil(..)-1, but this is because we start from 0
% calculate shift ranges: non_supported_intervals = [min_I, 255 - ([(quant_num_levels-1) * bin_size] + mid_val)] % we definetely don't want any bin value to be < min_I or > max_I
% the inclusive range covered between the extreme bin vals min2max_tight = (quant_num_levels - 1)*bin_size + 1
root_bin_vect = 0: bin_size: (quant_num_levels - 1)*bin_size
first_bin_vect = min_I: max_I - min2max_tight + 1 % ** in practice would have to use some logic to % get the smaller of the numbers in front
% initialize
% for qq = 1:length(first_bin_vect) % bin_val = first_bin_vect(qq) % shift_val = mid_val - bin_val %[5 : -16] % % shifted_I = (I + shift_val); % could contain - values or values >= % % positive_sI = shifted_I > 0; % truncated_sI = shifted_I .* positive_sI; % essentially mask the down-shifted I % % raw_quant_I = floor( truncated_sI ./ bin_size); % if max(raw_quant_I(:)) >= quant_num_levels % out_of_bounds_I_adjust = raw_quant_I >= quant_num_levels; % we know that in this case max(raw_quant_I)== % % so we use the nifty short cut. % % Else have to do a find and replace % % raw_quant_I = raw_quant_I - out_of_bounds_I_adjust; % end % % quant_I = bin_size * raw_quant_I + bin_val; % % max(quant_I(:)) % % CAE = sum(abs(I(:) - quant_I(:))) % cumilative absolute error % %my_MSE = sum((I(:) - quant_I(:)).^2)/length(I(:)) % matlab_MSE = MSE(I(:) - quant_I(:)) % if qq == 1 % min_MSE = matlab_MSE; % first_bin = bin_val;
shifted_I = (dithered_I + shift_val); % could contain - values or values >=
positive_sI = shifted_I > 0; truncated_sI = shifted_I .* positive_sI; % essentially mask the shifted I
raw_quant_I = floor( truncated_sI ./ bin_size); if max(raw_quant_I(:)) >= quant_num_levels out_of_bounds_I_adjust = raw_quant_I >= quant_num_levels; % we know that in this case max(raw_quant_I)== % so we use the nifty short cut. % Else have to do a find and replace
raw_quant_I = raw_quant_I - out_of_bounds_I_adjust; end
quant_dI = bin_size * raw_quant_I + bin_val;
matlab_MSE2 = MSE(I(:) - quant_dI(:))
figure, subplot(131), imshow(uint8(I)), title('original image') subplot(132), imshow(uint8(quant_I)), title('quantized image from part (a)'), xlabel('MSE = 41.809') subplot(133), imshow(uint8(quant_dI)), title(sprintf('dithered and then quantized image\nusing the same bins as in part (a)')) xlabel(sprintf('MSE = %.3f',matlab_MSE2))
(b)
I_thresh = I_green > 40; figure, subplot(221), imshow(I) subplot(222), imshow(I_thresh)
% 2. BWLABEL and identify the largest foreground object (but fill holes first) hole_filled_I_thresh = bwfill(I_thresh, 'holes')
[L, num_connected_objs] = bwlabel(hole_filled_I_thresh,4); % L is the LABELED img % num_connected_objs is FOREGROUND objects (0-bin is background) [count, bins] = hist(L(:), 0:num_connected_objs); [max_val, pos] = max(count(2:end)) % pos ends up being the right val because the % first count value is for the zero bin
Largest_FG_Obj = L==pos subplot(223), imshow(Largest_FG_Obj)
% 3. At this point we have to define "Area of the island" % i. Total Area of island (including water bodies etc, such as lakes) area_1 = max_val * (30/(1e3))^ xlabel(sprintf... ('Approximate "total area" of the island is %.2f square kilometers', area_1))
% ii. Area of land part alone of the island (mask I_thresh with % Largest_FG_Obj)
masked_island = I_thresh & Largest_FG_Obj; subplot(224), imshow(masked_island) [count_2, bins_2] = hist(masked_island(:), [0 1]); area_2 = count_2(2) * (30/(1e3))^ xlabel(sprintf... ('Approximate "land area" of the island is %.2f square kilometers', area_2))
Problem 1.3:
The above figure shows the sequences involved in the convolution. x [ n 1 , n 2 ] is shown as solid dots; x[ n 1 - m 1 , n 2 - m 2 ] is shown as a shaded region. Both sequences are wedge- shaped and the apex of x[ n 1 - m 1 , n 2 - m 2 ] is at the point [ n 1 , n 2 ]. If n 2 n 1 ≥ 0, the region of overlap is a parallelogram; it is zero otherwise. The parallelogram has a horizontal extent of n 1 +1 samples and a vertical extent of n 2 - n 1 +1 samples. Therefore,
y[ n 1 , n 2 ] = ( n 1 +1)( n 2 - n 1 +1), if n 2 ≥ n 1 ≥ 0 0, otherwise.
