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数学代写|离散数学代写discrete math代考|Equivalence Relations and Counting

如果你也在 怎样代写离散数学discrete math这个学科遇到相关的难题,请随时右上角联系我们的24/7代写客服。离散数学discrete math是研究可以被认为是 “离散”(类似于离散变量,与自然数集有偏射)而不是 “连续”(类似于连续函数)的数学结构。离散数学研究的对象包括整数、图形和逻辑中的语句。相比之下,离散数学不包括 “连续数学 “中的课题,如实数、微积分或欧几里得几何。离散对象通常可以用整数来列举;更正式地说,离散数学被定性为处理可数集的数学分支(有限集或与自然数具有相同心数的集)。然而,”离散数学 “这一术语并没有确切的定义。

离散数学discrete math的研究在二十世纪后半叶有所增加,部分原因是数字计算机的发展,它以 “离散 “的步骤操作,并以 “离散 “的比特存储数据。离散数学的概念和符号在研究和描述计算机科学分支的对象和问题时非常有用,如计算机算法、编程语言、密码学、自动定理证明和软件开发。反过来说,计算机实现在将离散数学的思想应用于现实世界的问题中也很重要,例如在运筹学中。

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my-assignmentexpert™ 为您的留学生涯保驾护航 在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在离散数学discrete math代写方面经验极为丰富,各种离散数学discrete math相关的作业也就用不着 说。

我们提供的离散数学discrete math及其相关学科的代写,服务范围广, 其中包括但不限于:

数学代写|离散数学代写discrete math代考|Equivalence Relations and Counting

数学代写|离散数学代写discrete math代考|Equivalence Relations and Equivalence Classes

In counting k-element subsets of an n-element set, we counted the number of lists of k distinct elements, getting n \underline{k}=n ! /(n-k)$ ! lists. Then we observed that two lists are equivalent as sets if I get one by rearranging (or “permuting”) the other. This divides the lists up into classes, called equivalence classes, all of size $k !$. The product principle told us that if $m$ is the number of such lists, then $m k !=n ! /(n-k)$ ! and we got our formula for $m$ by dividing. In a way it seems as if the proof does not account for the symmetry of the expression \frac{n !}{k !(n-k) !}. The symmetry comes of course from the fact that choosing a k element subset is equivalent to choosing the $n-k$ element subset of elements we don’t want. A principle that helps in learning and understanding mathematics is that if we have a mathematical result that shows a certain symmetry, it helps our understanding to find a proof that reflects this symmetry. We saw that the binomial coefficient $\left(\begin{array}{l}n \ k\end{array}\right)$ also counts the number of ways to label $n$ objects, say with the labels “in” and “out,” so that we have k “ins” and therefore n-k “outs.” For each labelling, the $k$ objects that get the label “in” are in our subset. Here is a new proof that the number of labellings is $n ! / k !(n-k) !$ that explains the symmetry.

Suppose we have $m$ ways to assign $k$ labels of one type and $n-k$ labels of a second type to $n$ elements. Let us think about making a list from such a labelling by listing first the objects with the label of type 1 and then the objects with the label of type 2 . We can mix the k elements labeled 1 among themselves, and we can mix the n-k labeled 2 among themselves, giving us k !(n-k) ! lists consisting of first the elements with label 1 and then the elements with label 2 . Every list of our $n$ objects arises from some labelling in this way. Therefore, by the product principle, m k !(n-k) ! is the number of lists we can form with n objects, namely n !. This gives us $m k !(n-k) !=n !$, and division gives us our original formula for $m$. With this idea in hand, we could now easily attack labellings with three (or more) labels, and explain why the product in the denominator of the formula for the number of labellings with three labels is what it is.
We can think of the process we described above as dividing the set of all lists of $n$ elements into classes of lists that are mutually equivalent for the purposes of labeling with two labels. Two lists of the $n$ objects are equivalent for defining labellings if we get one from the other by mixing the first $k$ elements among themselves and mixing the last $n-k$ elements among themselves. Relating objects we want to count to sets of lists (so that each object corresponds to an set of equivalent lists) is a technique we can use to solve a wide variety of counting problems.

数学代写|离散数学代写discrete math代考|Equivalence class counting

We can think of Exercise 1.3-2 as a problem of counting equivalence classes by thinking of writing down a list of the four people who are going to sit at the table, starting at some fixed point at the table and going around to the right. Then each list is equivalent to three other lists that we get by shifting everyone around to the right. (Why not shift to the left also?) Now this divides the set of all lists up into sets of four lists each, and every list is in a set. The sets are disjoint, because if a list were in two different sets, it would be equivalent to everything in both sets, and so it would have more than three right shifts distinct from it and each other. Thus we have divided the set of all lists of the four names into equivalence classes each of size four, and so by Theorem $1.3 .1$ we have $4 ! / 4=3 !=6$ seating arrangements.

Exercise $1.3-3$ is similar in many ways to Exercise 1.3-2, but there is one significant difference. We can visualize the problem as one of dividing lists of $n$ distinct beads up into equivalence classes, but turning a polygon or necklace over corresponds to reversing the order of the list. Any combination of rotations and reversals corresponds to natural geometric operations on the polygon, so an equivalence class consists of everything we can get from a given list by rotations and reversals. Note that if you rotate the list $x_{1}, x_{2}, \ldots, x_{n}$ through one place to get $x_{2}, x_{3}, \ldots, x_{n}, x_{1}$ and then reverse, you get $x_{1}, x_{n}, x_{n-1}, \ldots, x_{3}, x_{2}$, which is the same result we would get if we first reversed the list $x_{1}, x_{2}, \ldots, x_{n}$ and then rotated it through $n-1$ places. Thus a rotation followed by a reversal is the same as a reversal followed by some other rotation. This means that if we arbitrarily combine rotations and reversals, we won’t get any lists other than the ones we get from rotating the original list in all possible ways and then reversing all our results. Thus since there are $n$ results of rotations (including rotating through $n$, or 0 , positions) and each one can be flipped, there are $2 n$ lists per equivalence class. Since there are $n$ ! lists, Theorem 1.3.1 says there are $(n-1) ! / 2$ bead arrangements.

In Exercise 1.3-4, if all the checkers were distinguishable from each other (say they were numbered, for example), there would be $n-1+k$ objects to list, so we would have $(n+k-1)$ ! lists. Two lists would be equivalent if we mixed the red checkers among themselves and we mixed the black checkers among themselves. There are $k$ ! ways to mix the red checkers among themselves, and $n-1$ ways to mix the black checkers among themselves. Thus there are $k !(n-1)$ ! lists per equivalence class. Then by Theorem 1.3.1, there are
$$
\frac{(n+k-1) !}{k !(n-1) !}=\left(\begin{array}{c}
n+k-1 \
k
\end{array}\right)
$$
equivalence classes, so there are the same number of $k$-element multisets chosen from an $n$-element set.

数学代写|离散数学代写discrete math代考|Equivalence Relations and Counting

离散数学代写

数学代写|离散数学代写DISCRETE MATH代考|EQUIVALENCE RELATIONS AND EQUIVALENCE CLASSES

在计算 n 元素集合的 k 元素子集时,我们计算了 k 个不同元素的列表数,得到 \underline{k}=n ! /(n-k)$ ! lists. Then we observed that two lists are equivalent as sets if I get one by rearranging (or “permuting”) the other. This divides the lists up into classes, called equivalence classes, all of size $k !$. The product principle told us that if $m$ is the number of such lists, then $m k !=n ! /(n-k)$ ! and we got our formula for $m$ by dividing. In a way it seems as if the proof does not account for the symmetry of the expression \frac{n !}{k !(n-k) !}. The symmetry comes of course from the fact that choosing a k element subset is equivalent to choosing the $n-k$ element subset of elements we don’t want. A principle that helps in learning and understanding mathematics is that if we have a mathematical result that shows a certain symmetry, it helps our understanding to find a proof that reflects this symmetry. We saw that the binomial coefficient $\left(\begin{array}{l}n \ k\end{array}\right)$ also counts the number of ways to label $n$ objects, say with the labels “in” and “out,” so that we have k “ins” and therefore n-k “outs.” For each labelling, the $k$ objects that get the label “in” are in our subset. Here is a new proof that the number of labellings is $n ! / k !(n-k) !$解释了对称性。

假设我们有米分配方式ķ一种类型的标签和n−ķ第二种类型的标签n元素。让我们考虑通过首先列出具有类型 1 的标签的对象,然后列出具有类型 2 的标签的对象,从这样的标签中创建一个列表。我们可以混合标记为 1 的 k 个元素,也可以混合标记为 2 的 nk 个元素,得到 k !n−ķ!列表首先包含标签为 1 的元素,然后是标签为 2 的元素。我们的每一个清单n对象以这种方式从一些标签中产生。因此,根据乘积原理,mk !n−ķ!是我们可以用 n 个对象组成的列表的数量,即 n!。这给了我们米ķ!(n−ķ)!=n!, 除法给了我们原来的公式米. 有了这个想法,我们现在可以轻松地使用三个这r米这r和标签,并解释为什么用三个标签的标签数量公式的分母中的产品是什么。
我们可以认为我们上面描述的过程是划分所有列表的集合n将元素放入相互等效的列表类中,以便用两个标签进行标记。两个名单n如果我们通过混合第一个对象从另一个对象中获取一个对象,则对象等同于定义标签ķ元素之间并混合最后n−ķ元素之间。将我们要计数的对象与列表集相关联s这吨H一种吨和一种CH这bj和C吨C这rr和sp这nds吨这一种ns和吨这F和q在一世在一种l和n吨l一世s吨s是一种我们可以用来解决各种计数问题的技术。

数学代写|离散数学代写DISCRETE MATH代考|EQUIVALENCE CLASS COUNTING

我们可以把练习 1.3-2 看作是一个计算等价类的问题,可以考虑写下四个人的名单,这些人将坐在桌子旁,从桌子上的某个固定点开始,向右转。然后每个列表等效于我们通过将每个人向右移动获得的其他三个列表。在H是n这吨sH一世F吨吨这吨H和l和F吨一种ls这?现在这将所有列表的集合分成四个列表的集合,每个列表都在一个集合中。这些集合是不相交的,因为如果一个列表在两个不同的集合中,它将等同于两个集合中的所有内容,因此它将具有三个以上相互不同的右移。因此,我们将四个名称的所有列表的集合划分为等价类,每个等价类的大小为 4,因此由定理1.3.1我们有4!/4=3!=6座位安排。

锻炼1.3−3在许多方面与练习 1.3-2 相似,但有一个显着差异。我们可以将问题可视化为划分列表之一n不同的珠子变成等价类,但是翻转多边形或项链对应于颠倒列表的顺序。旋转和反转的任何组合都对应于多边形上的自然几何运算,因此等价类包含我们可以通过旋转和反转从给定列表中获得的所有内容。请注意,如果您旋转列表X1,X2,…,Xn通过一个地方得到X2,X3,…,Xn,X1然后反转,你得到X1,Xn,Xn−1,…,X3,X2,这与我们首先反转列表时得到的结果相同X1,X2,…,Xn然后旋转它n−1地方。因此,旋转之后的反转与反转之后的一些其他旋转相同。这意味着,如果我们任意组合旋转和反转,除了通过以所有可能的方式旋转原始列表然后反转所有结果得到的列表之外,我们将不会得到任何列表。因此,由于有n旋转的结果一世nCl在d一世nGr这吨一种吨一世nG吨Hr这在GH$n$,这r0,p这s一世吨一世这ns而且每一个都可以翻转,有2n每个等价类的列表。既然有n!列出,定理 1.3.1 说有(n−1)!/2珠安排。

在练习 1.3-4 中,如果所有的棋子都可以相互区分s一种是吨H和是在和r和n在米b和r和d,F这r和X一种米pl和, 那里将会是n−1+ķ要列出的对象,所以我们会有(n+ķ−1)!列表。如果我们将红色棋子混合在一起,将黑色棋子混合在一起,则两个列表将是等效的。有ķ!混合红色跳棋的方法,以及n−1混合黑色棋子的方法。因此有ķ!(n−1)!每个等价类的列表。然后由定理 1.3.1,有
(n+ķ−1)!ķ!(n−1)!=(n+ķ−1 ķ)
等价类,所以有相同数量的ķ- 元素多重集从一个n-元素集。

数学代写|离散数学代写discrete math代考

数学代写|离散数学代写discrete math代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。

微观经济学代写

微观经济学是主流经济学的一个分支,研究个人和企业在做出有关稀缺资源分配的决策时的行为以及这些个人和企业之间的相互作用。my-assignmentexpert™ 为您的留学生涯保驾护航 在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在图论代写Graph Theory代写方面经验极为丰富,各种图论代写Graph Theory相关的作业也就用不着 说。

线性代数代写

线性代数是数学的一个分支,涉及线性方程,如:线性图,如:以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。

博弈论代写

现代博弈论始于约翰-冯-诺伊曼(John von Neumann)提出的两人零和博弈中的混合策略均衡的观点及其证明。冯-诺依曼的原始证明使用了关于连续映射到紧凑凸集的布劳威尔定点定理,这成为博弈论和数学经济学的标准方法。在他的论文之后,1944年,他与奥斯卡-莫根斯特恩(Oskar Morgenstern)共同撰写了《游戏和经济行为理论》一书,该书考虑了几个参与者的合作游戏。这本书的第二版提供了预期效用的公理理论,使数理统计学家和经济学家能够处理不确定性下的决策。

微积分代写

微积分,最初被称为无穷小微积分或 “无穷小的微积分”,是对连续变化的数学研究,就像几何学是对形状的研究,而代数是对算术运算的概括研究一样。

它有两个主要分支,微分和积分;微分涉及瞬时变化率和曲线的斜率,而积分涉及数量的累积,以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互联系,它们利用了无限序列和无限级数收敛到一个明确定义的极限的基本概念 。

计量经济学代写

什么是计量经济学?
计量经济学是统计学和数学模型的定量应用,使用数据来发展理论或测试经济学中的现有假设,并根据历史数据预测未来趋势。它对现实世界的数据进行统计试验,然后将结果与被测试的理论进行比较和对比。

根据你是对测试现有理论感兴趣,还是对利用现有数据在这些观察的基础上提出新的假设感兴趣,计量经济学可以细分为两大类:理论和应用。那些经常从事这种实践的人通常被称为计量经济学家。

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。

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