BOOK CORRECTIONS, CLARIFICATIONS, AND CORRECTIONS TO PROBLEM SOLUTIONS

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1 Digital Image Processing, nd Ed. Gonzalez and Woods Prentice Hall 00 BOOK CORRECTIONS, CLARIFICATIONS, AND CORRECTIONS TO PROBLEM SOLUTIONS NOTE: Depending on the country in which you purchase the book, your copy may have most of the following errors corrected. To tell which printing of the book you have, look on the page (near the beginning of the book) that contains the Library of Congress Catalog card. Toward the bottom of that page, you will see a line that says: Printed in the United States of America. Below that line there is a series of numbers, starting with 0 on the left. The last number on the right is the printing of your book (e.g., if it is a, then you have the first printing of the book, a indicates that you have the second printing, and so on). The first column of each row in the following table indicates the printing(s) to which the corrections shown in that row apply. The following errata histories apply to U.S. printings. Printing histories of international editions do not always corresponds to the history of the books printed in the U.S. December, 008 BOOK CORRECTIONS Printing Page Reads Should Read vii, Page numbers for Chapter 5, 5, 7..., 44, 45, 3, 7,...,30, 3 (Entries are off by 4) xviii, nd paragraph, line 4 works Works (for MathWorks) 7, Fig , Caption for Fig..0 () Another... (b) Another , 6th line from the top... is simply the smallest number is simply the largest number... 7, Problem.0, 9 th line... 8 pixels each of bits each of , 5th line... between 9 and between 8 and , Fig. 3.6, (intersection of the s t axis and the dashed line) 93, st line, nd paragraph... continuos continuous... 94, last line... continuos continuous... 97, st line below Eq. (3.3-4)... continuos continuous... 00, Fig. 3.0, caption Photos Phobos -3 3, line below Eq. (3.4-5)... and σ η ( x, y) are and -3 8, Eq. (3.5-3) R = wz + wz + wz 9 9 R = wz + wz + + wz 9 9 σ η( x, y ) are...(remove bar over η ) -3 4, last line in Example removal of additive salt and removal of salt and... f f f f -3 8, Eqs. (3.7-) and (3.7-3),, x y x y -3 34, caption for Fig (a) Laplacian... (b) Laplacian... =3 37, caption for Fig Optical image of... (a) Optical image of... 43, Problem 3.(c) (c) What is the smallest value of s that (c) What is the smallest value of E that... 54, Eq. (4.-8) Fuv (, ) = [ R( xy, ) + I( xy, )] / Fuv (, ) = [ R( uv, ) + I( uv, )] /

2 Page of December, , Eq. (4.-9) j ( u) Fu ( ) = Fu ( ) e φ j ( u) Fu ( ) = Fu ( ) e φ -3 57, second line... miscroscope microscope... 67, 7 th line from bottom... origin center , caption for Fig. 4.6 using a GHPF of order with using a GHPF with , Eq. (4.4-5)... = ( ju) n F( u )... = ( j π u) n F( u) -6 85, Eq. (4.4-6)... = ( ju) F( u, v) + ( jv) F( u, v)... = ( j πu) F( u, v) + ( j πv) F( u, v) = ( u ) F( u, v) = 4 π ( u ) F( u, v) -6 85, Eq. (4.4-7)... = ( u ) F( u, v)... = 4 π ( u ) F( u, v) -6 85, Eq. (4.4-8)... = ( u )... = 4 π ( u ) 87, 3 rd parag., 3 rd line Huv (, ) = [( u M/ )... Huv (, ) = + [( u M/ )... 87, Eq. (4.4-3)...{ (( u M / ) +... }...{ + (( u M /) +... } -5 88, Fig. 4.8 Fig. 4.8(d) (printed incorrectly) Should look like Fig. 3.40(d), pg , 0th line from bottom is not as sharp as Fig. 4.43(d). The is not as sharp as Fig. 3.43(d). The 04, nd parag, line filter function to be inverse filter function to be multiplied by ( ) u+ v, transformed inverse transformed -3 09, line below Eq. (4.6-39) uk W =... ux W =... (replace k with x in the superscript) K -3 09, Eq. (4.6-40) ux ux... f ( xw ) K... f ( xw ) K (remove the in the subscript) -6 0, Table 4., 0th entry... ( ju) n F( u, v)... ( j π u) n F( u, v) ( jx) n f ( x, y)... ( j π x) n f( x, y) , Table 4., th entry... = ( u ) F( u, v)... = 4 π ( u ) F( u, v), Table 4., 4th entry from ( x y ) bottom, left side of equation A πσe ( x y ) Aπσ e π σ [ δ( u+ u, Table 4., nd 0, v0) + δ( u u0, v v0) ] entry from [ δ( u + Mu0, v + Nv0) + δ( u Mu0, v Nv0) ] bottom, right side of equation (Note: This answer in the book is correct, but it is for continuous (This answer is for discrete quantities.) quantities.) j [ δ( u+ u0, v0) δ( u u0, v v0) ], Table 4., last entry, right j [ δ( u+ Mu0, v+ Nv0) δ( u Mu0, v Nv0) ] side of equation (Note: This answer in the book is correct, but it is for continuous (This answer is for discrete quantities.) quantities.) 5, Prob. 4.4, second equation. A ( x y ) πσe ( x y ) Aπσ e π σ (See Corrections to Problem Solutions toward the end of this document). (Hint: Treat the variables as continuous (These closed-form expressions are applicable -6 5, Prob. 4.4, last line to simplify manipulation.) only to continuous variables.) -3 36, 3rd line from the top... d = ( mn ) / d = mn , last line of Level A: Else output z xy. Else output z med. -3 4, 4th parag, 6th line z xy z med 44, Example 5.6, 3 rd and 4 th sentences It is not difficult to show... pair of impulses. One impulse is located... mirror images of that point. K It is not difficult to show that the Fourier transform of a sine consists of two impulses that are mirror images of each other about the origin of the transform. Their locations are given in Table , Eq. (5.9-3) η = MN σ η m η η = MN σ η m η + (Note: Two corrections) 79, Problem D sine function D continuous sine function , Prob. 5., nd equation. πσ( u ) e π σ ( u ) πσ ( u ) e π σ ( u )

3 Page 3 of December, , Prob is linear and position invariant and show that is linear and position invariant and that the noise and image are uncorrelated. Show that... All 84, st parag., 0th line In ALL printings, should read:... the achromatic notion... All 86, nd parag., 3rd line In ALL printings, should read:... the achromatic notion... 94, 5 th line from top... CCCCCC... CCFFFF , four lines from bottom... point is the HSI point in the HSI... 99, Eq. (6.-7) G = (R + B) G = 3I (R + B) 300, Eq. (6.-) B = (R + G) B = 3I (R + G) 300, Eq. (6.-5) R = (G + B) R = 3I (G + B) -3 30, nd line from top... Fig. 6.5(d) Fig. 6.5(c) , 3 rd parag, nd line... those events our beyond... those events are beyond -3 33, nd parag., st line In this section be begin... In this section we begin , nd line below Eq. (6.5-8) (a,a,, a n)... ( a, a,, a n) , figure caption, nd... does not always alter the image hues sentence... does not alter the image hues. significantly. Variables inside the summation and on Change (x, y) to (s, t) inside the summations -3 38, Eqs. (6.6-) and (6.6-) the lower limits are shown as (x, y). and in the lower limits. S xy remains as is , caption for Fig, 6.5 abcd icons ab icons (stacked vertically) 344, References, nd line Gegenfurtne Gegenfurtner , line below Eq. (7.-3) determinate determinant , third line from top determinate determinant , nd line, nd paragraph... biorthogonality of the biorthogonality (defined below) of the , Table 7., nd row, nd col. H ( z) H ( z ) = H ( z) H ( z ) = , st... transform itself is both separable line and symmetric and can be transform can be , Eq. (7.-4) T = HFH T = HFH T 36, Eq. (7.-7), Second row, st column... we note that Fig. 7.8(a) is not the Haar -3 36, st line, last paragraph... we note that Fig. 7.8(a) bears... transform of Eqs. (7.-4)-(7.-6), but it bears , nd line of equation... (k l) (l k)... 39, 3 lines above Example functions behave vectors are used , last and next-to-last lines in the st paragraph , Eq. (7.6-5) in Fig Note the evenly... full packet decompositions. VJ = VJ WJ, D WJ, AA WJ, AD in Fig Note that the naturally ordered outputs of the filter bank in Fig. 7.30(a) have been reordered based on frequency content in Fig. 7.30(b) (see Problem 7.5 for more frequency ordered wavelets). VJ = VJ WJ, A WJ, DA WJ, DD , Figures 7.30(b) and 7.3 Figures 7.30(b) and 7.3 are incorrect. See corrected figures later in this document , nd line from top string of A s and D s... string of A s, H s, V s, and D s , Problem 7.9(b), line... transform is F = H - TH -, where T transform is F = H T TH, where T is the is the Haar transform of Haar transform of F , Problem 7.9(b), line and H - denotes the matrix inverse of and H T is the matrix inverse of H. Show that Haar transformation matrix H. Find H - T = H , Problem 7.9(b), line 3 H - for Haar transformation matrix H and use it to compute the inverse , Problem 7.5, last line sions for determining them , nd paragraph, 4 th line r n K J and use it to compute the inverse sions for determining them. Then order the basis functions according to frequency content and explain the results. r n K J , nd paragraph, 7 th line... equivalent to block encoding equivalent to encoding , nd paragraph, 8 th line... binary code words bit code words , 3rd paragraph, nd line formation units per symbol)... formation units per source symbol)...

4 Page 4 of December, , 3rd paragraph, 5 th line... is log( ϕ / r) and... is r logϕ (information units per code symbol) and... ϕ , Eq. (8.3-) R = log R = log ϕ r r , Eq. (8.4-8) fˆ n ( x, y ) = f ˆ( x, y ) = 47, last paragraph, line frequently used transformation frequently used transformations , st matrix, row 6 col , nd matrix, row 6 col last line in the example , Fig. 8.46, bottom, left -3 a subimage HH ( u, v ) a LH ( u, v ) -6 55, Problem 8.8. Q = [ ]. Q = [ ] T Square in (a) shows no size. The scale -3 56, Fig. 9.6 of (c) and (e) is too large. See corrected image later in this document. 53, Fig. 9. (caption) (Note that (d) Dilation of the opening. (e) Dilation of the opening. changes are to labels only). (e) Closing of the opening. (f) Closing of the opening. 54, 6th line from bottom. This -3 change corresponds to the revised Fig. 9.. See comment on pg... B B This change corresponds to the revised Fig. 9.. See comment Figure 9.(k)... Fig. 9.(l)... Figure 9.(l)... Fig. 9.(m)... on pg , Fig. 9. Figure 9. is incorrect. See corrected figure later in this document. 545, Fig. 9.4, st column, nd row (remove shading from some of the squares) -6 55, Fig. 9.7 Figure 9.7 is incorrect. See corrected figure later in this document , st parag, th line in such as way... in such a way... 57, nd parag, last line for the sake continuity... for the sake of continuity , lines from bottom of page would be reversed for an adge would be reversed for an edge , Fig. 0.7, 3rd column Incorrect signs for nd derivative. The signs of the nd derivatives in the 3rd column of the figure should be as in Fig , Eq. (0-3-3)... + σ σ σ σ , nd line from top continuos continuous -3 6, 6th line from bottom (e) PR ( i Rj) = FALSE... (e) PR ( i Rj) = FALSE for any adjacent regions Ri and R j. -3 6, 3rd line from top theresholding thresholding (e) PRi ( Rj) = FALSE for any adjacent -4 6, 6th line from bottom (e) PR ( i Rj) = FALSE for i j. regions R i and R j. -4 6, 5th line from bottom... P(R i ).... R i... Change the sub i to sub k for clarity. All 6, Fig. 0.45(d) See below for corrected figure , nd to last line in Ex the peak at in... the peak at u = 4 in 640, figure for Prob ( x a)... Replace all x s with z s , end of 0th line from bottom C x 687, 3 rd line from bottom and going into st paragraph of pg Interest in polygonal representation of boundaries (see Section..) is considerable because of its usefulness and simplicity. Typical work being done in this area T C y. For a detailed discussion and algorithm to compute minimum-perimeter polygons (Section..) see Sklansky et al. [97]. Typical work on polygonal approximations , Eq. (.-6)... ( m m ) ( m m )... ( m m ) T ( m + m ) , nd line below.3. i j i j.. denoted aa,, a n and bb,, b m.. i j i j.. denoted aa a n and bb b m , line below Eq. (.3-5)... of the symbols in a... of the corresponding symbols in a

5 Page 5 of December, , 9 lines from bottom... the measure R for five... the measure R for six , Index Gray code, 446 Gary code, , 8 th reference from bottom Shi Shih Change to: 38, Fig. 6.3(e) (The two graphs in the bottom, right). (Note that only the labels are changed the graphs remain the same). Change to: , Fig. 7.30(b) Change to: , Fig. 7.3

6 Page 6 of December, 008 Change to: -3 56, Fig. 9.6 Change to: f b A -6 55, Fig. 9.7 x W x W max{[ fx ( ) bs ( x) ]} max{[ fx ( ) b(0 x) ]} A b(0 x) bs ( x) 0 s s W/ W/ s

7 Page 7 of December, 008 Page 54 In addition to correcting some errors, the figures below Part (a) in Fig. 9. are improved considerably if replaced with the figures shown on the right. Note that the figure captions will run now from (a) through (m), so add another box containing an m at the bottom, right of the little icons. Then, make the following minor changes in the text. -3 In page 54, Section 9.5.5: 3rd line from the bottom of the last paragraph, change the superscript on B from 4 to 6. nd line from the bottom of that same paragraph: Change 9.(k) to 9.(l). Change 9.(l) to 9.(m). In page 54, caption of Fig. 9., nd line from bottom:... first element again (there was no change for the next two elements). Change that portion of the nd line to:... first four elements again. Last line: Change (k) to (l) and (l) to (m). The figure shows the corrected version of Fig The correction is on the left part of Fig. 0.45(d), in the light gray segment toward the top right. All =================== CLARIFICATIONS ) It has been pointed out to us that the black backgrounds in a few images throughout the book have small white specs. This appears to be a totally random printing problem that affected different images. Generally, the specs are clearly not part of the original images, but there are some cases in

8 Page 8 of December, 008 which confusion car arise if the size of the specs are comparable to white dots that are supposed to be there. A good example is Fig. 4.3(c), which consists of 5 symmetrically placed small white dots. Random, small white specs have been known to cause confusion on images like this. ) Restating the line above Eq. (3.3-), page 05, in the following way might make it clearer that the objective is to compute the mean value of the gray levels of the pixels in the neighborhood S xy :... the mean gray-level value, m Sxy using the expression..., of the pixels contained the neighborhood S xy can be computed Also in page 05, replace the material starting at the second line from the bottom, beginning with The values of this constant... and ending on the fourth line on the top of page with k < k, with the following clarification: Keeping in mind that we are interested in enhancing dark areas, and using the standard deviation as a measure of contrast, we will consider a pixel at point ( x, y ) to be a candidate for enhancement only if the standard deviation of its neighborhood, σ S, is in the range kd xy G σ S kd xy G, where D G is the global standard deviation, and k and k are positive constants such that k < k. In other words, kd G and kd G are the lower and upper limits of local contrast that determine whether a pixel is a candidate for modification. For example, if we choose k <, then only pixels whose local contrast is less than the global contrast become candidates for enhancement. The principal function of the lower limit is to prevent constant areas (whose standard deviation is 0) from being modified. A pixel at ( x, y ) that meets... 3) Page Units of frequency. If f ( xyis, ) an intensity function, and x and y are spatial coordinates, then the frequency variables of the Fourier transform represent changes in intensity per units of spatial distance. The units of frequency then are the reciprocal of the units of x and y, as Equations (4.-5) and (4.-6) show. 4) This clarification is in the second paragraph of pg. 04. The spatial representation, hxy (, ), of the filter transfer function, H ( uv, ), is obtained by taking the inverse DFT of this function. We know from the discussion earlier in this section that periodicity causes four adjacent periods of a DFT to meet in the center of the frequency rectangle (see Fig. 4.34). We handled this difficulty by multiplying the function by ( ) x+ y before computing its forward transform. We must do exactly the same when computing hxy (, ) via the inverse DFT. Failure to do this will yield incorrect convolution results. Even if all filtering is done in the frequency domain, we still need to compute hxy (, ) because padding is done in the spatial domain. Thus, to pad H ( uv, ) we must: () multiply this function by ( ) u+ v ; () compute its inverse Fourier transform to obtain hxy (, ); (3) pad this function with zeros; and (4) compute the forward DFT of the padded function to obtain the final representation of H ( uv, ) that we can use in frequency domain filtering. Note that we do not multiply hxy (, ) by ( ) x+ y prior to computing the forward transform because doing so would undo the centering of hxy (, ). In other words, the convolution theorem tells us that convolution of hxy (, ) and f ( xy, ) is being performed automatically even when the filtering

9 Page 9 of December, 008 operations are done in the frequency domain. Multiplying hxy (, ) by ( ) x+ y before computing its forward DFT to obtain our padded H ( uv, ) would undo the original centering obtained when we multiplied H ( uv, ) by ( ) u+ v prior to computing its inverse DFT. 5) Page 35, Caption of Fig. 7.. Add the following clarifying statement to the end of the figure caption: Note that all operations, including filtering, downsampling, and upsampling, are two-dimensional. ===================

10 Page 0 of December, 008 CORRECTIONS TO PROBLEM SOLUTIONS PROBLEM 4.4: The second equation in the problem statement should be revised to ( x y ) hxy (, ) = Aπσ e π σ + ( u )/σ We want to show that the inverse Fourier transform of Ae is equal to this function. Rather than doing it as shown in the original problem solution, it will be clearer if we start with one variable and show that, if then Hu ( ) = e u /σ ( ) =I [ Hu ( )] u /σ jπux = π x σ hx e e du = πσe. We can express the integral in the preceding equation as We now make use of the identity e Inserting this identity into the preceding integral yields, u j4πσ ux σ hx ( ) = e du. ( π) x σ ( π) x σ e =. ( π) x σ 4 u j4 πσ ux ( π ) σ x σ hx ( ) = e e du ( π) x σ u jπσ x σ = e e du. Next we make the change of variables r = u jπσ x. Then dr = du and the preceding integral becomes ( π) x σ r σ hx ( ) = e e dr. Finally, we multiply and divide the right side of this equation by πσ and obtain ( π) x σ r σ hx ( ) = πσe e dr πσ. The expression inside the brackets is recognized as a Gaussian probability density function, whose value from to is. Therefore, x hx ( ) = πσe π σ. With this result as background, we are now ready to show that ( u )/σ hxy (, ) =I [ Ae ] π σ ( x + y ) = A πσ e. By substituting directly into the definition of the inverse Fourier transform we have: ( u )/σ j π( ux+ vy) h( x, y) = Ae e dudv u v ( + j πuy) ( + j πvy) σ σ = Ae du e dv. The integral inside the brackets is recognized from the previous discussion to be equal to A the preceding integral becomes πσ x e π σ. Then,

11 Page of December, 008 v ( + j π vy) π σ x σ hxy (, ) = A πσe e dv. We now recognize the remaining integral to be equal to result: πσ y e π σ π σ x π σ y hxy (, ) = ( A πσe )( πσe ) π σ ( x + y ) = A πσ e., from which we have the final PROBLEM 4.5(a): There is a discrepancy in sign between the problem statement and the solution. To be consistent with the problem statement, the solution should start with g( xy, ) = f( x+, y) f( xy, ) + f( xy, + ) f( xy, ). In practice, one sees both formulations used interchangeably since ultimately we use squares or absolute values, which are independent of the sign used in the definition of the derivative. PROBLEM 5.3: The images shown in the solution of this problem are incorrect. They should be identical to the ones in PROBLEM 5.5. PROBLEM 6.6(a): 5 th line: Delete the words the entire. 6 th line: Change 55/360 to 360/55. PROBLEM 7.9. (a) Equation (7.-8) defines the Haar transformation matrix as Thus, using Eq. (7.-4), we get H =. T T = HFH = = (b) First, compute H = a c b d such that a c b d = 0 0. Solving this matrix equation yields T H = = = H H.

12 Page of December, 008 Thus, F T = H TH = = PROBLEM 7.5. The table is completed as follows: The functions are determined using Eqs. (7.-8) and (7.-8) with the Haar scaling and wavelet vectors from Examples 7.5 and 7.6: ϕ ()= x ϕ( x)+ ϕ( x ) ψ ()= x ϕ( x) ϕ( x ) To order the wavelet functions in frequency, count the number of transitions that are made by each function. For example, V 0 has the fewest transitions (i.e., only ) and correspondingly lowest frequency content, while W,AA has the most (i.e., 9) transitions and correspondingly highest frequency content. From top to bottom in the figure, there are, 3, 5, 4, 9, 8, 6, and 7 transitions, respectively. Thus, the frequency ordered subspaces are (from low to high frequency) V 0, W 0, W,D, W, A, W,DA, W,DD, W,AD, and W,AA.