RLE = [ ; ], with compression ratio (CR) = 4/8. RLE actually increases the size of the compressed image.

Transcription

1 MP/BME 574 Application Solutions. (2 pts) a) From first principles in class, we expect the entropy of the checkerboard image to be since this is the bit depth of the image and the frequency of each value ( and ) is identical. Recall also that a flat image (i.e. constant valued at all pixels) will have entropy since there is not information content by definition. A = [ ; ]; I = repmat(a,32,32); size(i) entropy(i) Flat = ones(64); entropy(flat) b) We can evaluate the compression ratio for run length encoding (RLE) by evaluating the fundamental repeated matrix R = [ ; ]. The RLE result from this 2 X 2 matrix is: RLE = [ ; ], with compression ratio (CR) = 4/8. RLE actually increases the size of the compressed image. For a 64 X 64, the R block is repeated 32 times so the CR will be the same for the larger image as it is for the block. c) There is no compression gained by variable length (Huffman) encoding either because the bit depth of the information in the checkerboard is bit, which is equal to the entropy. d) However, significant compression can be attained by using a block encoding scheme, which is natural to this problem since we know the checkerboard image is generated from this approach. Using the R block: R = [ ; ] contains 4 bits which is repeated in a 32 X 32 array. Therefore, the compression ratio is: CR = (64 X 64 X bit)/(2 X 5bits + 4 bits) = 496/4 = (2 pts) Repeating problem for the square test image. a) Now entropy is expected to be substantially lower because the bit depth of the image is much larger than the expected information content. The entropy is.6 bits/pixel, much lower than the bit depth of the image, which is 7 bits. Recall that the uniform intensity of the values within the square is 28.

2 MP/BME 574 Application Solutions Square = zeros(64); Insert = 28.*ones(8); Square(29:36,29:36)=Insert; entropy(square).6 b) For RLE, the rows -28 and are entirely zeros. If we allow breaks in the RLE at each image row, then each of these rows collapses to [ 64]. The compressed image by rows is then: [ 64] X 28 = 7 bits X 28 = 96 bits [ 28; 28 8; 28] X 8 = 22 bits X 8 = 76 bits [ 64] X 28 = 7 bits X 28 = 96 bits Total = 568 bits CR = (64 X 64 X 7)/568 =.5 (c) There is a bit depth of 7 for this image and an entropy of.2, so variable length encoding should significantly compress this image. Binary Code Probability of pixel intensity, i Huffman Code bits/pixel.563 Entropy =.6 bit/pixel CR = 7. Note that the primary compression for this image comes from assigning the 7 bit value for the squares to a bit number. Any further savings is limited because the bit depth cannot be less than bit/pixel. In a more complex image the CR could potentially be much higher. d) The square image is less amenable to a block encoding scheme than the checkerboard. I opted to use the example I showed in class for the block encoding of a binary border. The problem is simplified by assuming a priori knowledge that the image contains a square. The fundamental boundary block is then: R = [ ; ]; Now the square boundary can be encoded using 4 replicates of R combined with 2 reflections and rotation. The block itself is 4 bits.

3 MP/BME 574 Application Solutions Reflection Reflection Total = 4bits X 4 + reflection + rotation + reflection + 6 bits X 2 (matrix size) = 3 bits. CR = 496/3 = (5 pts) These are all lossless compression methods that exploit coding and interpixel redundancies. 4. (25 pts) a. Test Image Discrete Cosine Transform, Log Scale 24 DC or Zero Frequency Term k xmax k ymax DFT k-space Image, Log Scale k ymax k xmax DC or Zero Frequency Term k xmax k ymax

4 MP/BME 574 Application Solutions b. DFT Thresholded to : Inverse DFT Thresholded to : DCT Thresholded to : Inverse DCT Thresholded to : c. The primary artifacts are blurring and ringing. The blurring results from loss of detail coefficients representing higher frequency harmonics in the images. The ringing us due to the sharp truncation applied when thresholding the values in transform space. Note the slightly crisper depiction of structures for the thresholded discrete cosine transform coefficients relative to the DFT result. d. The relatively clear depiction of the basic structures in the images in part c is a manifestation of psychovisual redundancy. Lossy compression methods such as JPEG exploit this type of information redundancy. DCT and DFT Thresholding Code: I = phantom; FI = fftshift(fft2(fftshift(i))); figure;imagesc(abs(fi));axis('image');colormap('gray') figure;imagesc(log(abs(fi)));axis('image');colormap('gray') FI_thresh = FI(find(abs(FI)>.*max(max(FI)))); figure;imagesc(log(abs(fi_thresh)));axis('image');colormap('gray') [i,j] =find(abs(fi)>.274*max(max(fi))) size(j) FI_thresh = FI(i,j); FI_thresh = zeros(256); FI_thresh(i,j) = FI(i,j);

5 MP/BME 574 Application Solutions figure;imagesc(log(abs(fi_thresh)));axis('image');colormap('gray') figure;imagesc(abs(fftshift(ifft2(fftshift(fi_thresh)))));axis('image');colormap('gray') CI_thresh= zeros(256); CI = dct2(fftshift(i)); figure;imagesc(log(abs(ci)));axis('image');colormap('gray') [i,j] =find(abs(ci)>.224*max(max(ci))) size(j) CI_thresh(i,j) = CI(i,j); figure;imagesc(log(abs(ci_thresh)));axis('image');colormap('gray') CI_thresh = zeros(256); CI_thresh(i,j) = CI(i,j); figure;imagesc(log(abs(ci_thresh)));axis('image');colormap('gray') figure;imagesc(abs(idct2(ci_thresh)));axis('image');colormap('gray') 5. (3 pts) Problem 5: The key to the : compressed images is as follows: A = Discrete Fourier Transform (DFT), B = Wavelet encoding, C = JPEG encoding, D = Original Uncompressed Image, and E Discrete Cosine Transform (DCT). The major evidence allowing one to determine the compression type used includes: blurring indicated by the line profiles at left and visible block encoding in the magnified Image C. Clearly A and E are either DFT or DCT due to the blurring in the line profiles at left. There is less blurring in Image E suggesting it is DCT because this transform is known to more efficiently encode spatial frequency information (needs fewer coefficients). This leaves images B-D as Wavelet, Original or JPEG. Since C shows block encoding, it is JPEG. B also shows compression artifacts, so it is wavelet and D is the original. 3 Line Profiles for Test Images A-F Location of profile cuts across boundary and lesion in upper left corner of image A B C D E A B C D E

6 MP/BME 574 Application Solutions Display Code Used for Problem 5: [Px,Py,P,xi,yi] = improfile; Pa = improfile(a,xi,yi); Pc = improfile(c,xi,yi); Pd = improfile(d,xi,yi); Pe = improfile(e,xi,yi); figure;plot(x,pa,x,pb,x,pc(:8,,),x,pd,x,pe)

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