《数字图像处理》(冈萨雷斯)第三版课后星标习题详解

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"《数字图像处理》第三版 (冈萨雷斯)课后习题答案，由Rafael C. Gonzalez和Richard E. Woods编写，电子工业出版社出版，包含所有星标标记的习题详细解答，可从官方网站下载。" 本文档是《数字图像处理》第二版的补充材料，专门提供了书中所有带星号习题的详细解决方案。这些答案不仅对学习者深入理解教材内容至关重要，而且可以帮助检验自我学习成果，巩固理论知识。问题2.1涉及到的是图像在视网膜上的成像原理。通过相似三角形的原理，可以计算出视网膜上对应点的直径x。例如，如果一个点在视网膜上的直径d为2单位，而该点在图像传感器（假设为模拟人眼的视网膜）上的尺寸为0.014单位，那么比例关系为d/x=2/0.014，解得x=0.07d。这展示了图像缩小的比例关系。在讨论中提到了视网膜可以类比为一个方形的传感器阵列，大约有337,000个元素，这意味着阵列大小约为580x580。考虑到传感器之间的等间距，1.5毫米长的线上会有580个元素和579个空隙。因此，每个元素和每个空隙的大小是1.5毫米除以580，即每个元素和空隙的尺寸。这个问题的解答揭示了如何将物理尺寸与图像传感器的像素尺寸联系起来，并强调了实际图像处理中几何比例和分辨率的概念。在数字图像处理中，这样的计算对于理解和设计图像采集系统、分辨率分析以及图像缩放算法至关重要。此外，书中的习题通常涵盖图像的获取、表示、变换、分析和增强等多个方面，包括颜色模型、灰度变换、滤波、边缘检测、特征提取等核心知识点。通过解决这些习题，读者能够深入理解图像处理的基本原理和操作，从而具备处理和分析数字图像的能力。《数字图像处理》第三版的课后习题答案是深入学习数字图像处理的宝贵资源，它提供了理论与实践相结合的机会，有助于提升读者在图像处理领域的专业知识和技能。

12 Chapter 3 Solutions (Students)

used for both negative and positive differences, in which case we have a band [¡T; T ]

in which all pixels of d(x; y) must fall in order for f(x; y) to be declared acceptable.

The second major approach is simply to sum all the pixels in jd(x; y)j and compare the

sum against a threshold S. Note that the absolute value needs to be used to avoid errors

cancelling out. This is a much cruder test, so we will concentrate on the ®rst approach.

There are three fundamental factors that need tight control for difference-based inspec-

tion to work: (1) proper registration, (2) controlled illumination, and (3) noise levels

that are low enough so that difference values are not affected appreciably by variations

due to noise. The ®rst condition basically addresses the requirement that comparisons

be made between corresponding pixels. Two images can be identical, but if they are

displaced with respect to each other, comparing the differences between them makes

no sense. Often, special markings are manufactured into the product for mechanical or

image-based alignment

Controlled illumination (note that |illumination} is not limited to visible light) obviously

is important because changes in illumination can affect dramatically the values in a

difference image. One approach often used in conjunction with illumination control is

intensity scaling based on actual conditions. For example, the products could have one

or more small patches of a tightly controlled color, and the intensity (and perhaps even

color) of each pixels in the entire image would be modi®ed based on the actual versus

expected intensity and/or color of the patches in the image being processed.

Finally, the noise content of a difference image needs to be low enough so that it does

not materially affect comparisons between the golden and input images. Good signal

strength goes a long way toward reducing the effects of noise. Another (sometimes

complementary) approach is to implement image processing techniques (e.g., image

averaging) to reduce noise.

Obviously there are a number if variations of the basic theme just described. For exam-

ple, additional intelligence in the form of tests that are more sophisticated than pixel-by-

pixel threshold comparisons can be implemented. A technique often used in this regard

is to subdivide the golden image into different regions and perform different (usually

more than one) tests in each of the regions, based on expected region content.

Problem 3.17

(a) Consider a 3 £3 mask ®rst. Since all the coef®cients are 1 (we are ignoring the 1/9

Problem 3.19 13

scale factor), the net effect of the lowpass ®lter operation is to add all the gray levels of

pixels under the mask. Initially, it takes 8 additions to produce the response of the mask.

However, when the mask moves one pixel location to the right, it picks up only one new

column. The new response can be computed as

new

= R

old

¡C

+ C

where C

is the sum of pixels under the ®rst column of the mask before it was moved,

and C

is the similar sum in the column it picked up after it moved. This is the basic

box-®lter or moving-average equation. For a 3 £3 mask it takes 2 additions to get C

was already computed). To this we add one subtraction and one addition to get

new

. Thus, a total of 4 arithmetic operations are needed to update the response after

one move. This is a recursive procedure for moving from left to right along one row of

the image. When we get to the end of a row, we move down one pixel (the nature of the

computation is the same) and continue the scan in the opposite direction.

For a mask of size n £ n, (n ¡ 1) additions are needed to obtain C

, plus the single

subtraction and addition needed to obtain R

new

, which gives a total of (n + 1) arith-

metic operations after each move. A brute-force implementation would require n

¡ 1

additions after each move.

Problem 3.19

(a) There are n

points in an n £ n median ®lter mask. Since n is odd, the median

value, ³, is such that there are (n

¡ 1)=2 points with values less than or equal to ³

and the same number with values greater than or equal to ³. However, since the area

A (number of points) in the cluster is less than one half n

, and A and n are integers,

it follows that A is always less than or equal to (n

¡1)=2. Thus, even in the extreme

case when all cluster points are encompassed by the ®lter mask, there are not enough

points in the cluster for any of them to be equal to the value of the median (remember,

we are assuming that all cluster points are lighter or darker than the background points).

Therefore, if the center point in the mask is a cluster point, it will be set to the median

value, which is a background shade, and thus it will be |eliminated} from the cluster.

This conclusion obviously applies to the less extreme case when the number of cluster

points encompassed by the mask is less than the maximum size of the cluster.

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