### 机器视觉代写|图像处理作业代写Image Processing代考|DISTANCE TRANSFORM

statistics-lab™ 为您的留学生涯保驾护航 在代写图像处理Image Processing方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写图像处理Image Processing代写方面经验极为丰富，各种代写图像处理Image Processing相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 机器视觉代写|图像处理作业代写Image Processing代考|DISTANCE TRANSFORM

Distance transform is an important tool in image processing, and it is normally only applied to binary images that consist of object and background pixels. A distance transform of a binary image specifies the distance from every pixel to the nearest background pixel. In other words, the distance transform converts a binary image into a grayscale image where each object pixel has a value corresponding to the minimum distance from the background. The resulting grayscale image is a so-called distance map.

Assume $f$ is a binary image, in which the pixels with a value of ‘ 0 ‘ indicate the background while the pixels with a value of ‘ 1 ‘ indicate the object. Let $B=$ ${p \mid f(p)=0}$ be the set of background pixels and $O={p \mid f(p)=1}$ be the set of object pixels. The distance transform of a binary image $f, D(p)$, can be given by [39]:
$$D(p)= \begin{cases}0, & \text { if } p \in B \ \min _{q \in B} d(p, q), & \text { if } p \in O\end{cases}$$
where function $d$ is a distance function or metric which is to determine the distance between pixels.

For pixels $p, q$, and $r$ in an image, a distance function $d$ satisfies the following three criteria [128]:

1. Positive definite: $d(p, q) \geq 0(d(p, q)=0$ iff $p=q)$
2. Symmetric: $d(p, q)=d(q, p)$
3. Triangular: $d(p, r) \leq d(p, q)+d(q, r)$There are several types of distance metrics in image processing. The three most important ones are: Euclidean, city-block, and chessboard.

## 机器视觉代写|图像处理作业代写Image Processing代考|PERFORMANCE OF THE DISTANCE METRICS

Figure 2.14 Effects of different distance transforms.
denoted by the operator ‘ $*$ ‘, is defined as:
\begin{aligned} h(x, y) &=\omega(x, y) * f(x, y) \ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \omega(u, v) f(x-u, y-v) \mathrm{d} u \mathrm{~d} v \end{aligned}
In image processing, where an image is represented by a set of pixels, convolution is a local operation that replaces each pixel in an image by a linear combination of its neighbors. The impulse response $\omega(x, y)$ is then referred to as a convolution kernel, and the convolution becomes the calculation of the sum of products of the kernel coefficients with the intensity values in the region encompassed by the kernel. The convolution of a kernel $\omega(x, y)$ of size $m \times n$ with an image $f(x, y)$ is given by:
\begin{aligned} h(x, y) &=\omega(x, y) * f(x, y) \ &=\sum_{s=-\frac{1}{2}}^{\frac{1}{2}} \sum_{t=-\frac{4}{2}}^{\frac{n}{2}} \omega(s, t) f(x-s, y-t) \end{aligned}
For each pixel $(x, y)$ in the image, the convolution value $h(x, y)$ is the weighted sum of the pixels in the neighborhood about $(x, y)$, where the individual weights are the corresponding coefficients in the convolution kernel. This procedure involves translating the convolution kernel to pixel $(x, y)$ in the image, multiplying each pixel in

the neighborhood by a corresponding coefficient in the convolution kernel, and summing the multiplications to obtain the response at each pixel $(x, y)$. Figure $2.15$ gives an example of convolution of an image with a $3 \times 3$ kernel. In this example, the response of the kernel at the center point $(x, y)$ of the $3 \times 3$ image neighborhood is given by:
\begin{aligned} h(x, y)=& \omega(-1,-1) f(x-1, y-1)+\omega(-1,0) f(x-1, y) \ &+\omega(-1,1) f(x-1, y+1)+\omega(0,-1) f(x, y-1) \ &+\omega(0,0) f(x, y)+\omega(0,1) f(x, y+1)+\omega(1,-1) f(x+1, y-1) \ &+\omega(1,0) f(x+1, y)+\omega(1,1) f(x+1, y+1) \end{aligned}

## 机器视觉代写|图像处理作业代写Image Processing代考|SET AND LOGICAL OPERATIONS

Since a binary image is a matrix containing object pixels of value 1 and background pixels of value 0 , it can simply be represented as the set of those coordinate vectors

$(x, y)$ of the pixels that have value of 1 in the binary image, given by:
$$G={(x, y) \mid g(x, y)=1}$$
where $(x, y)$ are pairs of spatial coordinates, $g(x, y)$ is the pixel value ( 0 or 1$)$ at $(x, y)$, and $G$ represents the set of image pixels describing the object of interest. All other image pixels are assigned to the background.

Let $\mathbb{Z}$ be the set of integers. Let the elements of a binary image be represented by a set $A \subseteq \mathbb{Z} \times \mathbb{Z}$, whose elements are 2-dimensional vectors of the form $(x, y)$, which are spatial coordinates. If a set contains no elements, it is called an empty set or a null set, denoted by $\varnothing$. If $\omega=(x, y)$ is an element of $A$, then it is written as:
$\omega \in A$
otherwise, it is written as:
$\omega \notin A$
If every element of a set $A$ is also an element of a set $B$, then $A$ is said to be a subset of $B$ and written as:
$$A \subseteq B$$
A set $B$ of pixel coordinates $\omega$ that satisfy a particular condition is written as:
$$B={\omega \mid \text { condition }}$$
The universe set, $\mathbb{U}$, is the set of all elements in a given application. In image processing, the universe is typically defined as the rectangle containing all the pixels in an image.

The complement (or inverse) of $A$, denoted as $A^{c}$, is the set of all elements of $U$ that do not belong to set $A$, given by:
$$A^{c}={\omega \mid \omega \notin A}=\mathbb{U}-A$$
The complement of the binary image $A$ is the binary image that exchanges black and white, that is, 0 -valued pixels set to 1 -valued and 1 -valued pixels set to 0 -valued.
The union of two sets $A$ and $B$, denoted as $A \cup B$, is the set of all elements that belong to either $A, B$, or both, given by:
$$A \cup B={\omega \in A \text { or } \omega \in B}$$
The union of two binary images $A$ and $B$, is a binary image in which the pixels’ values are 1 if the corresponding input pixels’ values are 1 in $A$ or in $B$.

Similarly, the intersection of two sets $A$ and $B$, denoted as $A \cap B$, is the set of all elements that belong to both $A$ and $B$, given by:
$$A \cap B={\omega \in A \text { and } \omega \in B}$$
The intersection of two binary images $A$ and $B$ is a binary image where the pixels’ values are 1 if the corresponding input pixels’ values are 1 in both $A$ and $B$.

## 机器视觉代写|图像处理作业代写Image Processing代考|DISTANCE TRANSFORM

D(p)={0, 如果 p∈乙 分钟q∈乙d(p,q), 如果 p∈这

1. 正定：d(p,q)≥0(d(p,q)=0当且当p=q)
2. 对称：d(p,q)=d(q,p)
3. 三角形：d(p,r)≤d(p,q)+d(q,r)图像处理中有几种类型的距离度量。最重要的三个是：欧几里得、城市街区和棋盘。

## 机器视觉代写|图像处理作业代写Image Processing代考|PERFORMANCE OF THE DISTANCE METRICS

H(X,是)=ω(X,是)∗F(X,是) =∫−∞∞∫−∞∞ω(在,在)F(X−在,是−在)d在 d在

H(X,是)=ω(X,是)∗F(X,是) =∑s=−1212∑吨=−42n2ω(s,吨)F(X−s,是−吨)

H(X,是)=ω(−1,−1)F(X−1,是−1)+ω(−1,0)F(X−1,是) +ω(−1,1)F(X−1,是+1)+ω(0,−1)F(X,是−1) +ω(0,0)F(X,是)+ω(0,1)F(X,是+1)+ω(1,−1)F(X+1,是−1) +ω(1,0)F(X+1,是)+ω(1,1)F(X+1,是+1)

## 机器视觉代写|图像处理作业代写Image Processing代考|SET AND LOGICAL OPERATIONS

(X,是)二值图像中值为 1 的像素数，由下式给出：
G=(X,是)∣G(X,是)=1

ω∈一种

ω∉一种

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。