数字图像处理第三版课后习题完整解答

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"这是一份关于《数字图像处理》第三版的英文版课后习题解答，包含了所有标记星号问题的详细解决方案。由Rafael C. Gonzalez和Richard E. Woods编写，由Prentice Hall出版社出版。本书旨在帮助学生深入理解数字图像处理的概念和技术。" 在《数字图像处理》第三版中，本书提供了丰富的学习资源，特别是对于那些希望进一步掌握图像处理理论和应用的学生。解答手册包含了所有章节中带星号的问题，这些通常是较为关键或具有挑战性的练习，可以帮助学生巩固课程中的核心概念。 1. 引言 (Introduction) 1.2 项目 (Projects) 书中鼓励学生进行计算机项目实践，以深化理论知识。每个项目应包含以下部分： - 封面页：项目标题、项目编号、课程编号、学生姓名、截止日期、提交日期和不超过半页的摘要。 - 技术讨论：1到2页的技术性内容，解释项目的技术要点。 - 结果讨论：1到2页，分析项目结果及其意义。 - 结果展示：打印的图像结果，每张图片必须与结果讨论中的编号和标题对应。 - 附录：列出原创代码，重点是学生自己编写的函数和程序。通过这样的结构，学生不仅可以锻炼解决问题的能力，还能提升技术写作和表达能力。此外，书中的项目部分还强调了实际操作和结果分析的重要性，这对于培养学生的实践能力和批判性思维至关重要。数字图像处理是一门涵盖广泛领域的学科，包括图像获取、图像增强、图像复原、图像分析、特征提取、图像分类和识别等。本解答手册将帮助学生在这些领域建立起坚实的基础，同时也为他们提供了自我评估和提高的工具。学习数字图像处理不仅对计算机科学、电子工程和相关专业的学生有益，也适用于从事图像处理、机器视觉、人工智能和数据分析等领域工作的专业人士。这份详尽的解答集为独立学习者和教师都提供了宝贵的资源，有助于深化对图像处理技术的理解和应用。

diti

ons (a) and (b) stated in that section. However, we see from Fig. P3.9(b) that

the inverse transformation from s back to r is not single valued, as there are an

inﬁnite number of p ossible mappings from s =(L −1) /2backtor .Itisimpor-

tant to note that the reason the inverse transformation function turned out not

to be single valued is the gap in p

(r ) in

the interval [L/4, 3L/4].

Problem

.10

(b) I

none of the intensity levels r

, k = 1

,...,L −1, are 0, then T (r

) will

trictly monotonic. This implies a one-to-one mapping both ways, meaning that

both forward and inverse transformations will be single-valued.

Problem 3

.12

The v

alue of the histogram component corresponding to the kth intensity level

in a neighborhood is

for k = 1

,2,...,K −1, where n

is t

he number of pixels having intensity level r

n is

the total number of pixels in the neighborhood, and K is the total number

of possible intensity levels. Suppose that the neighborhood is moved one pixel

to the right (we are assuming rectangular neighborhoods). This deletes the left-

most column and introduces a new column on the r ight. The updated histogram

then becomes



−n

+ n

]

for k = 0

,1,...,K −1, where n

is the

number of occurrences of level r

on t

left column and n

is t

he similar quantity on the right column. The preceding

equation can be written also as



)=p

−n

]

for k = 0

,1,...,K −1. The same concept applies to other modes of neighborhood

motion:



)=p

−a

]

for k = 0

,1,...,K −1, where a

is t

he number of pixels with value r

in t

he neigh-

borhood area deleted by the move, and b

the corresponding number intro-

duced by the move.

CHAPTER 3

. PROBLEM SOLUTIONS

Problem 3

.13

The p

urpose of this simple problem is to make the student think of the meaning

of histograms and arrive at the conclusion that histograms carry no information

about spatial properties of images. Thus, the only time that the histogram of the

images formed by the operations shown in the problem statement can be de-

termined in terms of the original histograms is when one (both) of the images

is (are) constant. In (d) we have the additional requirement that none of the

pixels of g (x, y ) can be 0. Assume for convenience that the histograms are not

normalized, so that, for example, h

) is t

he number of pixels in f (x, y ) having

intensity level r

. A

ssume also that all the pixels in g (x, y ) have constant value c .

The pixels of both images are assumed to be positive. Finally, let u

denote

the

intensity levels of the pixels of the images formed by any of the arithmetic oper-

ations given in the problem statement. Under the preceding set of conditions,

the histograms are determined as follows:

(a) We obtain the histogram h

sum

) of

the sum by letting u

= r

+ c,an

dalso

sum

)=h

) for

all k . In other words, the values (height) of the compo-

nents of h

sum

the same as the components of h

ut their locations on the

intensity axis are shifted right by an amount c .

Problem 3

.15

(a) Co

nsider a 3 ×3 mask ﬁrst. Because all the coefﬁcients are 1 (we are ignoring

the 1/9 scale factor), the net effect of the lowpass ﬁlter operation is to add all the

intensity values 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

new

= R

old

−C

+ C

wher

e C

is t

he sum of pixels under the ﬁrst column of the mask before it was

moved, and C

is t

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

is the basic box-ﬁlter or moving-average equation. For a 3 ×3 mask it takes 2

additions to get C

was

lready computed). To this we add one subtraction

and one addition to get R

new

. T

hus, 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.

F or a mask of size n ×n, (n −1) additions are needed to obtain C

,pl

usthe

single subtraction and addition needed to obtain R

new

, w

hich gives a total of

(n + 1) ari

thmetic operations after each move. A brute-force implementation

would require n

−1 a

dditions after each move.

Problem 3

.16

(a) The

key to solving this problem is to recognize (1) that the convolution re-

sult at any location (x,y ) consists of centering the mask at that point and then

forming the sum of the products of the mask coefﬁcients with the corresponding

pixels in the image; and (2) that convolution of the mask with the entire image

results in every pixel in the image being visited only once by every element of

the mask (i.e., every pixel is multiplied once by every coefﬁcient of the mask).

Because the coefﬁcients of the mask sum to zero, this means that the sum of the

products of the coefﬁcients with the same pixel also sum to zero. Carrying out

this argument for every pixel in the image leads to the conclusion that the sum

of the elements of the convolution array also sum to zero.

Problem 3

.18

(a) Ther

e are n

poi

nts in an n × n median ﬁlter mask. Because n is odd, the

median value, ζ,issuchthatthereare(n

−1)/2

oints with values less than or

equal to ζ and the same number with values greater than or equal to ζ.How-

ever, because the area A (number of points) in the cluster is less than one half

,an

dA and n are integers, it follows that A is always less than or equal to

−1)/2.

Thus, even in the extreme case when all cluster points are encom-

passed by the ﬁlter 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 .

Problem 3

.19

(a) N

umerically sort the n

val

ues. The median is

ζ =[(n

+ 1)/2]-th l

argest value.

(b) Oncethevalueshavebeensortedonetime,wesimplydeletethevaluesin

the trailing edge of the neighborhood and inser t the values in the leading edge

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