Haar Transform CS 430 Denbigh Starkey

Transcription

1 Haar Transform CS Denbig Starkey. Background. Computing te transform. Restoring te original image from te transform 7. Producing te transform matrix 8 5. Using Haar for lossless compression 6. Using Haar for lossy compression

2 Background Te Haar transform as a rater messy looking definition, so I ll give te definition and will ten go carefully troug an example. Te transform is important for a couple of reasons: It leads to efficient image compression scemes, bot lossless and lossy, wic I ll describe later in tese notes, and it provides a lead-in to wavelets, wic I ll be covering in te next set of notes. I ll continue to assume te image is, were is a power of, and tat J = log. In te pyramid sceme we created a number of new images, leading to an increase of pixels wit an upper bound of a % increase. Wit Haar our transformed image will ave te same dimensions as te original, and so any compression savings are immediately gained.

3 Computing te Haar Transform Given an image F, te Haar transform will be computed as: T = HFH T. ote: Gonzalez incorrectly omits te transpose. were H contains te Haar basis functions. Te matrix H will be defined wit te structure: H = Te definition of tese k z blending functions first needs a side definition of two additional variables, p and q, for eac row k. Tey are defined using four rules: k = p + q p J if p =, ten q = or if p, ten q p Tese are used to define te k z functions as follows. z = p / q q if z < p p p / q q k z = pq z = if z <. p p oterwise I ll build te H matrix slowly using te matrix as an example, were = and so J =. First I ll get te a and b values tat we need for k =,, and. For all of tese cases te second rule for p and q states tat p is eiter or, since J =.

4 If k = = p + q, ten eiter p = and so q =, or p = and so q =. Te second option violates te fourt rule and so p = and q =. If k = = p + q, ten eiter p = and so q =, or p = and so q =. Te first option violates te tird rule and so p = and q =. If k = = p + q, ten eiter p = and so q =, or p = and so q =. Te first option violates te tird rule and so p = and q =. ow we need te matrix: H =. ote tat all of tese values are multiples of, wic is in tis case, so we can pull tis value outside te matrix. Te first row is easy, since every element is defined as being. Te second row as k =. As we saw above, p = and q =. So te definition says tat: z = z = if z < if z <. oterwise So =, =, =, and =. ow consider te tird row, wit k =. As we saw above, p = and q =. Te definition says tat:

5 5 z = z = < < oterwise z if z if. So =, =, =, and =. For k = we found tat p = and q =. So, using te definition, z = z = < < oterwise z if z if. So =, =, =, and =. Putting tis all togeter, we finally get: H =. ote tat Gonzalez and Woods ave one of tese values wrong. Te Haar transform is easier to develop since J =, and so p. I.e., p can only be zero. For k = tis gives = + q, and so q =. z = for all z, wic fixes te first row. z = z = < < oterwise z if z if. Te functions are evaluated at z = and z =, giving te matrix:

6 H =. Gonzalez and Woods, in Figure 7.8, sows te result of applying te H Haar discrete wavelet transform to an image, but doesn t explain a lot of concepts tat are going on in te figure. E.g., one migt get te impression from tat figure tat Haar will produce a reduced copy of te original image in te top left corner of te transformed image, but tis doesn t appen. As I ll discuss below, tis reduced image is actually scattered troug te transformed array, but Gonzalez and Woods ave reorganized te array structure to make it more visually intuitive. Tey also don t sow ow easy it is to take te transformed image and restore te original. I ll look at te restoration problem first, and ten describe wat is going on in Figure 7.8 and all oter transformed images. Finally I ll look at ow to use Haar in lossless and lossy compression scemes. 6

7 Restoring te Original Image from te Transform Look at te rows of H and H as vectors. If we take te dot product of any pair of tese row vectors ten te result is zero, wic means tat in H te four rows form ortogonal vectors in -space and in H tey form ortogonal vectors in -space. In addition te magnitude of eac row vector remembering to include te multiplier outside te matrix is equal to. Tecnically tis means tat tese and all oter Haar matrices are ortonormal, wic also means tat te inverse of eac Haar matrix is its transform. Te discrete Haar transform formula is, as mentioned above, T = HFH T. Because H is ortonormal tis can be rewritten as: and so T = HFH -, H - TH = F. Reusing te ortonormal property means tat we can restore F from T using: F = H T TH. 7

8 Producing te Transform Matrix As discussed above, te image sown in Figure 7.8 of Gonzales and Woods is not te Haar transform image, but as been rearranged for visual convenience. Tis rearrangement is very standard, but you sould know wat is being done to get te image. I ll use H in te discussion below, since leads to simpler descriptions, but te extension to, say, H and is trivial. First, remember tat: H =. We divide te matrix up into blocks, and apply te transform to eac block to get te block in te transformed image. E.g., if te block as a and b in its first row and c and d in its second row te transform will be H a c b d H T = a c b d oting tat te transpose of H is te same as H, wic is not true for oter Haar matrices. Tis gives te transformed block: a + b + c + d a + b c d a b + c d a b c + d. Look at wat is going on ere. Te top left is a lowpass or mean filter. Te top rigt is an average orizontal gradient, and te bottom left an average vertical gradient. Te bottom rigt gives diagonal curvature. Applying tis to an image of Peppy on te left, below, gives te result on te rigt, wic doesn t ave any of te structure sown in, say, Figure 7.8 of Gonzalez and Woods. However te cat is clearly recognizable for two reasons. One is tat every fourt pixel in te new image is an average of four pixels in te original, and te oter is tat te image also includes te vertical, orizontal, and diagonal curvature filter information. Te Peppy image was 5 5 wit 56 gray levels. Te transform produced values between -55 e.g., a = b =, c = d = 55 on te lower left and +5 all four variables = 55 on te top left. So te transformed image was 8

9 normalized to be between and 55 for display by adding 55 and dividing by. Peppy Peppy wit Haar Transform Te form in Gonzalez and Woods Figure 7.8 is te result of moving te pixels around. In tis case we ll get four tiles, eac 56 56, were te top left corner is all of te top left pixels in eac block, te top rigt is all of te top rigt pixels in eac block, etc. So we get te result sown below: Peppy wit tiled Haar Transform I ve done te color balancing differently ere. Te top left as an intensity range of to 5, so I ve alved tose intensities to get to 55. Te oter 9

10 tree tiles ave a range of -55 to +55, and so for tem I ve added 55 and ten alved. I.e., te adjustment is not consistent troug te image. Te relevant code from my program for getting te original transform and te tiled transform, assuming obvious definitions and wit SIZE #defined as 5 is: for i = ; i < SIZE; i = i + { for j = ; j < SIZE; j = j + { a = inarray[i][j]; b = inarray[i][j + ]; c = inarray[i + ][j]; d = inarray[i + ][j + ]; outarray[i][j] = a + b + c + d / ; outarray[i][j + ] = a - b + c - d / ; outarray[i + ][j] = a + b - c - d / ; outarray[i + ][j + ] = a - b - c + d / ; outto55[i][j] = outarray[i][j] + 55 / ; outto55[i][j + ] = outarray[i][j + ] + 55 / ; outto55[i + ][j] = outarray[i + ][j] + 55 / ; outto55[i + ][j + ] = outarray[i + ][j + ] + 55 / ; } } iby = i / ; jby = j / ; sizeby = SIZE / ; prettyout[iby][jby] = outarray[i][j] / ; prettyout[iby][jby + sizeby] = outarray[i][j + ] + 55 / ; prettyout[iby + sizeby][jby] = outarray[i + ][j] + 55 / ; prettyout[iby + sizeby][jby + sizeby] = outarray[i + ][j + ] + 55 / ; If I take te outarray values and use tem as input to te same program I get Peppy back, as expected.

11 Using Haar for Lossless Compression I ll look at lossless compression ere and ten I ll look at lossy compression in te next part of tese notes. For lossless compression I ll use H as te example again, altoug as we ll see te compression is muc better for te iger H transforms. Wit H we get four tiles using te tiled view. Te top left tile doesn t compress any better tan te input image, but te oter tree tiles are clustered very close to zero. E.g., for Peppy over 8% of te pixels in tese tree tiles are between - and +, leaving only 8% of te pixels wit absolute values between and 55. Even better, tere are only seven pixels wit absolute values greater tan 7, out of a total of 96,68 pixels. So using some kind of variable lengt coding like Huffman coding some uge gains can be acieved for te tree filter tiles. If we were te use a iger H transform like, say, H 8, ten te reduced tile in te top left will now only contain 6 of te pixels in te image, and so te 6 compression will be over te remaining 6 of te pixels. Tis means tat significant compression can be acieved wit Huffman coding or similar systems.

12 Using Haar for Lossy Compression Tis is were one can win spectacularly using Haar. Most of te output of te Haar transform is, as I ve discussed before, centered around zero. E.g., even for te Peppy image, wic as some extreme differences in intensity, for H about 6% of te pixels in te tree modified tiles are at zero, and 8.9% are between - and +. Since tese are representing relatively low cange areas, one can take a band of tem around zero and cange tem all to zero witout affecting image quality wen te original image is restored. E.g., for te image below I applied H to Peppy and ten canged all values in te tree filter tiles tat were between - and + to zero. Ten, wen I restored it I got te image sown. I ve looked carefully at tis image and te original Peppy side by side on my screen at 5 5 eac, and aven t been able to see any differences at all. Tis means tat using H wit tis cange over 8% of te values in te 75% of te image tat represents filters ave te same value zero, and so any image compression sceme like Huffman or run-lengt encoding will be able to make uge gains. If I were to use, say, H 8, instead, I d be able to get similar compression 6 savings, but now over 6 of te file.