Digital image processing gonzalez 2nd edition pdf free download

For Eq. They are not orthonormal, however, since they do not satisfy the requirements of column 3 in Table 7. Using Eq. Figure P7. Note the gap between '1;0 x and '1;1 x. When the index is large, the resemblance is strongu else it is weak. Thus, if a function is similar to itself at different scales, the resemblance index will be similar at different scales. As a result, we can say that the function whose CWT is shown is self-similarxlike a fractal signal.

The CWT is often easier to interpret because the built-in redundancy tends to reinforce traits of the function or image. For example, see the self-similarity of Problem 7. To determine C, use Eq. To check the result substitute these values into Eq. To construct the approximation pyramid that corresponds to the transform in Fig. Thus, the approximation pyramid would have 4 levels.

If the input is shifted, the transform changes. This is the case in the third transform shown. They are 2. Since their sum is less than the starting entropy of 2. Because the detail entropy is 0, no further decomposition of the detail is warranted. Thus, we perform another FWT iteration on the approximation to see if it should be decomposed again.

This process is then repeated until no further decompositions are called for. The resulting optimal tree is shown in Fig. That is, all gray levels are equally probable. Since all levels are equally probable, there is no advantage to assigning any particular gray level fewer bits than any other.

Thus, we assign each the fewest possible bits required to cover the 2n levels. Problem 8. The maximum run length would be 2n and thus require n bits for representation. The starting coordinate of each run also requires n bits since it may be arbitrarily located within the 2n pixel line. Since a run length of 0 can not occur and the run-length pair 0; 0 is used to signal the start of each new line - an additional 2n bits are required per line.

To achieve some level of compression, CR must be greater than 1. Table P8. Error 12 8 4 16 12 16 -4 16 13 8 5 25 13 16 -3 9 10 8 2 4 13 16 -3 9 57 56 1 1 54 48 6 36 Problem 8. The rms error is then 3. The squared signal value i.

The remaining values of Table 8. The resulting code words are , , and , respectively. The correctly decoded binary value is In a similar manner, the parity words for and are and , respectively. The decoded values are identical and are The matrix multiplication of Eq. Thus, the decrease in uncertainty is 0. Substituting 0. Thus, the binary erasure channel with a higher probability of error has a larger capacity to transfer information. Thus, this is the maximum possible information compression under this criterion.

The codes are complements of one another. They are constructed by following the Huffman procedure for three symbols of arbitrary probability. The number of bits used for each gray level, however, should be the same for all versions constructed. The construction of Code 2 in Table 8.

Step 2: Assign code words based on the ordered probabilities from right to left, as shown in Fig. Step 3: The codes associated with each gray level are read at the left of the diagram. Note that two Huffman shift codes are listed, one of which is the best. In generating these codes, the sum of probabilities 4 - 7 were used as the probability of the shift up symbol. The sum is 0. Thus, the two codes shown differ by the ordering of r0 and the shift symbol during the Huffman coding process.

Shift 1 H. Start by dividing the [0, 1 interval according to the symbol probabilities. This is shown in Table P8. The decoder immediately knows the message 0. To further see this, divide the interval [0. Proceeding like this, which is the same procedure used to code the message, we get eaii! The coding proceeds as shown in Table P8. Address Dict. For the encoded output in Example 8. The output is then read from the third column of the table to yield 39 39 39 39 39 39 39 39 where it is assumed that the decoder knows or is given the size of the image that was recieved.

Note that the dictionary is generated as the decoding is carried out. Entry 39 39 39 39 39 39 Problem 8. The transition at c must somehow be tied to a particular transition on the previous line. Note that there is a closer white to black transition on the previous line to the right of c, but how would the decoder know to use it instead of the one to the left.

Both are less than ec. The second relationship is a valid one. The coding mode sequence is pass, vertical 1 left , vertical directly below , horizontal distances 3 and 4 , and pass. The initial approximation of the X-ray must contain no more than this number of bits.

At the X-ray encoder, the X-ray can be JPEG compressed using a normalization array that yields about a compression. The resulting binary image will contain a 1 in every bit position at which the approximation differs from the original. To achieve an average error-free bit-plane compression of 1. A conceptual block diagram for both the encoder and decoder are given below. Likewise, using Eqs. Figure P8. A simple approach is to double the image resolution in both dimensions by interpolation see Section 2.

Then, the appropriate 6-connected points are picked out of the expanded array. The resolution of the new image will be the same as the original but the former will be slightly blurred due to interpolation. Figure P9. The black points are the original pixels and the white points are the new points created by interpolation. The squares are the image points picked for the hexagonal grid arrangement.

Ambiguities arise when there is more than one path that can be followed from one 6-connected pixel to another.

In one-pixel-thick, fully connected bound- aries, these multiple paths manifest themselves in the four basic patterns shown in Fig. The solution to the problem is to use the hit-or-miss transform to detect the patterns and then to change the center pixel to 0, thus eliminating the multiple paths. For example, consider the sequence of points shown in Fig. Thus, we would end up with different but of course, acceptable m-paths. Keep in mind that erosion is the set described by the origin of the structuring element, such that the structuring element is contained within the set being eroded.

The intersection of two convex sets is convex also. See Fig. Keep in mind that the digital sets in question are the larger black dots. The lines are shown for convenience in visualizing what the continuous sets would be.

In b the result of dilation is not convex because the center point is not in the set. In c we see that the lower right point is not connected to the others. In d , it is clear that the two inner points are not in the set.

The center of each structuring element is shown as a black dot. Solution a was obtained by eroding the original set shown dashed with the structuring element shown note that the origin is at the bottom, right. Solution b was obtained by eroding the original set with the tall rectangular structuring element shown.

Then dilated image was then eroded with a disk of half the diameter of the disk used for dilation. Problem 9. But, as shown in Problem 9. If the origin of B is contained in B, then the set of points describing the erosion is simply all the possible locations of the origin of B such that B z is contained in A.

Now, from Eq. Let C denote the erosion of A by B. It was already established that C is a subset of A. From the preceding discussion, we know also that C is a subset of the dilation of C by B. The next to last step in the preceding sequence follows from the fact that the closing of a set by another contains the original set [this is from Problem 9. From Problem 9. A disk structuring element of radius 11 was used.

This structuring element was just large enough to encompass all noise elements, as given in the problem statement. The images shown in Fig. We do not consider this an important issue because it is scale-dependent, and nothing is said in the problem statement about this. The next dilation should eliminate the internal noise components completely and further increase the size of the rectangle. This also would have been true if the objects and structuring elements were rectangular. However, a complete reconstruction, for instance, by dilating a rectangle that was partially eroded by a circle, would not be possible.

The resulting of applying the hit-or-miss transform would be a single point where the two Tzs were in perfect registration. The location of the point would be the same as the origin of the structuring element. Major differences are the lack of a complex conjugate in the hit-or-miss transform, and the fact that this transform produced a single nonzero binary value in this case, as op- posed to the multiple nonzero values produced by correlation of the two images.

Assuming that the shapes are processed one at a time, basic two-step approach for differentiating between the three shapes is as follows: Step 1. Apply an end-point detector to the object until convergence is achieved. If the result is not the empty set, the object is a Lake. Otherwise it is a Bay or a Line. Step 2. There are numerous ways to differentiate between a lake and a line. One of the simplest is to determine a line joining the two end points of the object.

For example, square 2,2 in Fig. There would be other 1zs introduced that would turn Fig. This situation requires additional preprocessing, as discussed below. These connected components are labels with a value different from 1 or 0. The remaining black points are interior to spheres.

The simplest approach is to separate the spheres by preprocessing. This approach works in this case because the objects are spherical, thus having small areas of contact. To handle the case of spheres touching the border of the image, we simply set all border point to black.

Note that the erosion of white areas makes the black areas interior to the spheres grow, so the possibility exists that such an area near the border of a sphere could grow into the background. This issue introduces further complications that the student may not have the tools to solve yet.

Create an image of the same size as the original, but consisting of all 0zs, call it B. Choose an arbitrary point labeled 1 in A, call it p1 , and apply the algorithm. When the algorithm converges, a connected component has been detected. Choose an arbitrary point labeled 1 in A1 , call it p2 , and repeat the procedure just given.

If there are K connected components in the original image, this procedure will result in an image consisting of all 0zs after K applications of the procedure just given. Image B will contain K labeled connected components.

In terms of the intervals given in the problem state- ment, this means that x and y must be in the closed interval x 2 [Bx1 ; Bx2 ] and y 2 [By1 ; By2 ]. To remove these spikes we perform an opening with a cylindri- cal structuring element of radius greater than Rmax , as shown in Fig. Note that the shape of the structuring element is matched to the known shape of the noise spikes. A structuring element like the one used in a but with radius slightly larger than 4Rmax will do the job.

Note in a and b that other parts of the image would be affected by this approach. The bigger Rmax , the bigger the structuring element that would be needed and, consequently, the greater the effect on the image as a whole. Call the result- ing set of border pixels B.

Apply the connected component algorithm. All connected components that contain elements from B are particles that have merged with the border of the image. Determine the area number of pixels of a single particleu denote the area by R. Eliminate from the image the particles that were merged with the border of the image.

Apply the connected component algo- rithm. Count the number of pixels in each component. It also is stated that we can ignore errors due to digitizing and positioning. This means that the imaging system has enough resolution so that artifacts will not be introduced as a result of digitization. The mechanical accuracy similarly tells us that no appreciable errors will be introduced as a result of positioning. This is important if we want to do matching without having to register the images.

Because we are interested in boundary defects, the method of choice is a backlighting system that will produce a binary image. We are assured from the problem statement that the illu- mination system has enough resolution so that we can ignore defects due to digitizing.

The next step is to specify a comparison scheme. The simplest way to match binary images is to AND one image with the complement of the other.

If the images are identical and perfectly registered the result of the AND operation will be all 0zs. Otherwise, there will be 1zs in the areas where the two images do not match.

Note that this requires that the images be of the same size and be registered, thus the assumption of the mechanical accuracy given in the problem statement. As noted, differences in the images will appear as regions of 1zs in the AND image. These we group into regions connected components by using the algorithm given in Section 9. The most stringent criterion is 0 connected components. This means a perfect match. More sophisticated criteria might involve measures like the shape of connected components and the relative locations with respect to each other.

These types of descriptors are studied in Chapter Each mask would yield a value of 0 when centered on a pixel of an unbroken 3-pixel segment oriented in the direction favored by that mask. Figure P End points are those points on a line which have only one 8-neighbor valued 1. Once all end points have been found, the D8 distance between all pairs of such end points gives the lengths of the various gaps.

This is a rudimentary solution, and numerous embellishments can be added to build intelligence into the process. For example, it is possible for end points of different, but closely adjacent, lines to be less than L pixels apart, and heuristic tests that attempt to sort out things like this are quite useful. Problem Assume that n is odd and keep in mind that an ideal step edge transition takes place between adjacent pixels.

Thus, we get a ramp edge, as expected. The objective is to show that the responses of the Sobel masks are indistinguishable for these four edges. That this is the case is evident from Table P The same line of reasoning applies to the Prewitt masks. Table P The idea with the one-dimensional mask is the same: We replace the value of a pixel by the response of the mask when it is centered on that pixel. Next, we pass the mask 2 3 1 6 7 4 1 5 1 through these results.

Next we apply the smoothing mask [1 2 1] to these results. The process to show equivalence for Gy is basically the same. Note, however, that the directions of the one-dimensional masks would be reversed in the sense that the differencing mask would be a column mask and the smoothing mask would be a row mask.

The numbers in brackets are values of [Gx ; Gy ]. The histogram follows directly from this table. The average value of the convolution can be obtained by evaluating the Fourier transform of this product at the origin of the frequency plane [see Eq.

But, it was shown in a that the average value of r2 h is zero, which means that its Fourier transform is zero at the origin. From this it follows that the value of the product of the two Fourier transforms is also zero, thus proving that the average value of the convolution of f with r2 h is zero.

Consider Eq. Here, it follows from Eq. The same zero result is obtained for Eq. The boundary of each connected component forms a closed path Prob- lem 2. The contours in Fig. Torre and T. Pattern Analysis and Machine Intell. Looking up this paper and becoming familiar with the mathematical underpinnings of edge detection is an excellent reading assignment for graduate students.

Substituting into Eq. Thus the re-ective adjacency. Each group is then processed further to determine if its points satisfy the criteria for a valid track: 1 each group must have at least pointsu and 2 it cannot have more than three gaps, each of which cannot be more than 10 pixels long see Problem These paths are as follows: 1 : 1; 1 1; 2!

Assume that p is to the right as the image is traversed from left to right. The possible paths are shown in Fig. The costs are detailed in Fig. The graph with the minimum-cost path shown dashed is shown in Fig.

Finally, the edge corresponding to the minimum-cost path is shown in Fig. When the images are blurred, the boundary points will give rise to a larger number of different values for the image on the right, so the histograms of the two blurred images will be different.

These values are summarized in Table P The histogram of the product is shown in Fig. If the number of samples is not large, convergence to a value at or near the mid point between the two means also requires that a clear valley exist between the two modes.

A unique feature of the book is its emphasis on showing how to enhance those tools by developing new code. This is important in image processing, an area that normally requires extensive experimental work in order to arrive at acceptable application solutions. Some Highlights: 1 This new edition is an extensive upgrade of the book.

Gonzalez received the B. He served as Chairman of the department from through He also founded Perceptics Corporation in and was its president until The last three years of this period were spent under a full-time employment contract with Westinghouse Corporation, who acquired the company in Under his direction, Perceptics became highly successful in image processing, computer vision, and laser disk storage technologies.

In its initial ten years, Perceptics introduced a series of innovative products, including: The world's first commercially-available computer vision system for automatically reading the license plate on moving vehicles; a series of large-scale image processing and archiving systems used by the U. Navy at six different manufacturing sites throughout the country to inspect the rocket motors of missiles in the Trident II Submarine Program; the market leading family of imaging boards for advanced Macintosh computers; and a line of trillion-byte laser disk products.

He is a frequent consultant to industry and government in the areas of pattern recognition, image processing, and machine learning. Brooks Distinguished Professor Award. Dougherty Award for Excellence in Engineering in See our User Agreement and Privacy Policy. See our Privacy Policy and User Agreement for details. Published on May 2,. The lowest-priced brand-new, unused, unopened, undamaged item in its original packaging where packaging is applicable. Packaging should be the same as what is found in a retail store, unless the item is handmade or was packaged by the manufacturer in non-retail packaging, such as an unprinted box or plastic bag.

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