标签： CS141

计算机代写|数字硬件系统代写Digital hardware systems代考|ECE2544

Posted on 2023年2月14日2023年2月14日 by statistics-lab

如果你也在怎样代写数字硬件系统Digital hardware systems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。

数字硬件是指与有线电视和卫星电视系统以及DVD和DVC播放器有关的集成接收器解码器或其他视频解码器，其中每一个都包含一个集成电路，该集成电路包含一个用于实施付费传输过程的装置。

statistics-lab™ 为您的留学生涯保驾护航在代写数字硬件系统Digital hardware systems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。我们的专家在代写数字硬件系统Digital hardware systems代写方面经验极为丰富，各种代写数字硬件系统Digital hardware systems相关的作业也就用不着说。

我们提供的数字硬件系统Digital hardware systems及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

计算机代写|数字硬件系统代写Digital hardware systems代考|ECE2544

计算机代写|数字硬件系统代写Digital hardware systems代考|Routing Architecture

The routing architecture of an FPLD determines a way in which the programmable switches and wiring segments are positioned to allow the programmable interconnection of logic cells. A routing architecture for an FPLD must meet two criteria: routability and speed. Routability refers to the capability of an FPLD to accommodate all the nets of a typical application, despite the fact that the wiring segments must be defined at the time the blank FPLD is made. Only switches connecting wiring segments can be programmed (customized) for a specific application, not the numbers, lengths or locations of the wiring segments themselves. The goal is to provide a sufficient number of wiring segments while not wasting chip area. It is also important that the routing of an application can be determined by an automated algorithm with minimal intervention.

Propagation delay through the routing is a major factor in FPLD performance. After routing an FPLD, the exact segments and switches used to establish the net are known and the delay from the driving output to each input can be computed. Any programmable switch (EPROM, pass-transistor, or antifuse) has a significant resistance and capacitance. Each time a signal passes through a programmable switch, another RC stage is added to the propagation delay. For a fixed R and C, the propagation delay mounts quadratically with the number of series RC stages. The use of a low resistance switch, such as antifuse, keeps the delay low and its distribution tight. Of equal significance is optimization of the routing architecture. Routing architectures of some commercial FPLD families are presented in this section.

In order to present commercial routing architectures, we will use the routing architecture model shown in Figure 1.17. First, a few definitions are introduced in order to form a unified viewpoint when considering routing architectures.

A wire segment is a wire unbroken by programmable switches. One or more switches may attach to a wire segment. Typically, one switch is attached to the each end of a wire segment. A track is a sequence of one or more wire segments in a line. A routing channel is a group of parallel tracks.

计算机代写|数字硬件系统代写Digital hardware systems代考|FPLD Applications

FPLDs have been used in a large number of applications, ranging from the simple ones replacing glue logic to those implementing new computing paradigms, that are not possible using other technologies. In this section we will list some of them, as to make a classification into some typical groups, and emphasize most important features of each group.

CPLDs are used in applications that can efficiently use wide fan-in of AND/OR gates and do not need a large number of flip-flops. Examples of such circuits are various kinds of finite state machines. On the other hand, FPGAs with a large number of flip-flops are better suited for the applications that need memory functions and complex data paths. Also, due to their easy reprogrammability they become an important element of prototyping digital systems designs. As such they enable emulation of entire complex systems, and in many cases also their final implementation. Finally, all FPGAs as the static RAM based circuits allow at least a minimum level of dynamic reconfigurability. While all of them allow full device reconfiguration by downloading another bitstream (configuration file), some of them also allow partial reconfiguration. The partial reconfiguration provides change of the function of a part of the device, while the remaining part operates without disruption of the system function.

In order to visualize the range of current and potential applications, we have to mention typical features of FPLDs in terms of their capacity and speed. Today the leading suppliers of FPLDs offer devices containing up to 500,000 equivalent (twoinput NAND) gates, with a perspective to quadruple this figure in the next two to three years. These devices are delivered in a number of configurations so that application designers have the choice to fit their designs into a device with minimal capacity. They also come in a range of speed grades and different packages with different number of input/output pins. The number of pins sometimes exceeds 600 . The speed of circuits implemented in FPLDs varies depending primarily on application and design approach. As an illustration, all major manufacturers offer devices that provide full compliance with 64-bit $66 \mathrm{MHz}$ PCI-bus requirements.

数字硬件系统代考

计算机代写|数字硬件系统代写Digital hardware systems代考|Routing Architecture

FPLD 的路由架构决定了可编程开关和布线段的定位方式，以实现逻辑单元的可编程互连。FPLD 的布线架构必须满足两个标准：可布线性和速度。可布线性是指 FPLD 适应典型应用的所有网络的能力，尽管必须在制作空白 FPLD 时定义布线段。只有连接接线段的开关才能针对特定应用进行编程（定制），而不是接线段本身的数量、长度或位置。目标是在不浪费芯片面积的情况下提供足够数量的布线段。同样重要的是，应用程序的路由可以由自动化算法以最少的干预来确定。

通过路由的传播延迟是影响 FPLD 性能的主要因素。在对 FPLD 进行布线后，用于建立网络的确切段和开关是已知的，并且可以计算从驱动输出到每个输入的延迟。任何可编程开关（EPROM、传输晶体管或反熔丝）都有很大的电阻和电容。每次信号通过可编程开关时，传播延迟都会增加另一个 RC 级。对于固定的 R 和 C，传播延迟与串联 RC 级的数量呈二次方关系。使用低电阻开关（例如反熔丝）可保持较低的延迟和紧凑的分布。同样重要的是路由架构的优化。本节介绍了一些商业 FPLD 系列的路由架构。

为了展示商业路由架构，我们将使用图1.17所示的路由架构模型。首先，介绍一些定义，以便在考虑路由体系结构时形成统一的观点。

线段是可编程开关未断开的线。一个或多个开关可以连接到电线段。通常，一个开关连接到线段的每一端。轨道是一条线上的一系列一个或多个线段。路由通道是一组平行的轨道。

计算机代写|数字硬件系统代写Digital hardware systems代考|FPLD Applications

FPLD 已用于大量应用，从替换胶合逻辑的简单应用到实现新计算范例的应用，这些都是使用其他技术无法实现的。在本节中，我们将列出其中的一些，以将其分类为一些典型的组，并强调每个组的最重要特征。

CPLD 用于可有效使用宽扇入 AND/OR 门且不需要大量触发器的应用。这种电路的例子是各种有限状态机。另一方面，具有大量触发器的 FPGA 更适合需要存储功能和复杂数据路径的应用。此外，由于它们易于重新编程，它们成为数字系统设计原型设计的重要元素。因此，它们可以模拟整个复杂系统，在许多情况下还可以模拟它们的最终实施。最后，作为基于静态 RAM 的电路的所有 FPGA 至少允许最低水平的动态可重构性。虽然它们都允许通过下载另一个比特流（配置文件）进行完整的设备重新配置，但其中一些还允许部分重新配置。

为了可视化当前和潜在应用的范围，我们不得不提及 FPLD 在容量和速度方面的典型特征。如今，领先的 FPLD 供应商提供包含多达 500,000 个等效（双输入 NAND）门的器件，并有望在未来两到三年内将这一数字翻两番。这些设备以多种配置交付，因此应用程序设计人员可以选择将他们的设计融入具有最小容量的设备中。它们还具有一系列速度等级和具有不同数量输入/输出引脚的不同封装。引脚数有时会超过 600 个。在 FPLD 中实现的电路速度主要取决于应用和设计方法。例如，所有主要制造商都提供完全符合 64 位的设备66米H和PCI 总线要求。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

术语广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。

有限元方法代写

有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。

有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。

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。

随机分析代写

随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。

时间序列分析代写

随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。

回归分析代写

多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。

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，工具箱允许您学习和应用专业技术。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。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

计算机代写|数字硬件系统代写Digital hardware systems代考|CSE320

Posted on 2023年2月14日2023年2月14日 by statistics-lab

如果你也在怎样代写数字硬件系统Digital hardware systems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。

我们提供的数字硬件系统Digital hardware systems及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

计算机代写|数字硬件系统代写Digital hardware systems代考|CSE320

计算机代写|数字硬件系统代写Digital hardware systems代考|Antifuse Programming Technology

An antifuse is an electrically programmable two-terminal device. It irreversibly changes from high resistance to low resistance when a programming voltage (in excess of normal signal levels) is applied across its terminals. Antifuses offer several unique features for FPGAs, most notably a relatively low on-resistance of 100-600 Ohms and a small size. The layout area of an antifuse cell is generally smaller than the pitch of the metal lines it connects; it is about the same size as a via connecting metal lines in an MPLD. When high voltage (11 to 20 Volts) is applied across its terminals, the antifuse will “blow” and create a low resistance link. This link is permanent. Antifuses are built either using an Oxygen-Nitrogen-Oxygen (ONO) dielectric between an $\mathrm{N}+$ diffusion and polysilicon, or amorphous silicon between metal layers or between polysilicon and the first layer of metal.

Programming an antifuse requires extra circuitry to deliver the high programming voltage and a relatively high current of $5 \mathrm{~mA}$ or more. This is done through large transistors to provide addressing to each antifuse.

Antifuses are normally “off” devices. Only a small fraction of the total that need to be turned on must be programmed (about $2 \%$ for a typical application). So, other things being equal, programming is faster with antifuses than with “normally on” devices.

Antifuse reliability must be considered for both the unprogrammed and programmed states. Time dependent dielectric breakdown (TDDB) reliability over 40 years is an important consideration. It is equally important that the resistance of a programmed antifuse remains low during the life of the part. Analysis of ONO dielectrics shows that they do not increase the resistance with time. Additionally, the parasitic capacitance of an unprogrammed amorphous antifuse is significantly lower than for other programming technologies.

计算机代写|数字硬件系统代写Digital hardware systems代考|Logic Cell Architecture

In this section we present a survey of commercial FPLD logic cell architectures in use today, including their combinational and sequential portions. FPLD logic cells differ both in size and implementation capability. A two transistor logic cell can only implement a small size inverter, while the look-up table logic cells can implement any logic function of several input variables and is significantly larger. To capture these differences we usually classify logic blocks by their granularity.

Since granularity can be defined in various ways (as the number of Boolean functions that the logic block can implement, the number of two-input AND gates, total number of transistors, etc.), we choose to classify commercial blocks into just two categories: fine-grain and coarse-grain.

Fine-grain logic cells resemble MPLD basic cells. The most fine grain logic cell would be identical to a basic cell of an MPLD and would consist of few transistors that can be programmably interconnected.

The FPGA from Crosspoint Solutions uses a single transistor pair in the logic cell. In addition to the transistor pair tiles, as depicted in Figure 1.9, the cross-point FPGA has a second type of logic cell, called a RAM logic tile, that is tuned for the implementation of random access memory, but can also be used to build other logic functions.

A second example of a fine-grain FPGA architecture is the FPGA from Plessey. Here the basic cell is a two-input NAND gate as illustrated in Figure 1.10. Logic is formed in the usual way by connecting the NAND gates to implement the desired function. If the latch is not needed, then the configuration memory is set to make the latch permanently transparent.

Several other commercial FPGAs employ fine-grain logic cells. The main advantage of using fine-grain logic cells is that the usable cells are fully utilized. This is because it is easier to use small logic gates efficiently and the logic synthesis techniques for such cells are very similar to those for conventional MPGAs (MaskProgrammable Gate Arrays) and standard cells.

数字硬件系统代考

计算机代写|数字硬件系统代写Digital hardware systems代考|Antifuse Programming Technology

反熔丝是一种电可编程的两端设备。当在其端子上施加编程电压（超过正常信号电平）时，它不可逆地从高电阻变为低电阻。反熔丝为 FPGA 提供了几个独特的特性，最显着的是 100-600 欧姆的相对较低的导通电阻和小尺寸。反熔丝单元的布局面积通常小于其连接的金属线的间距；它的大小与 MPLD 中连接金属线的过孔大致相同。当在其端子上施加高压（11 至 20 伏）时，反熔丝将“熔断”并形成低电阻连接。这个链接是永久的。反熔丝是使用氧-氮-氧 (ONO) 电介质在否+金属层之间或多晶硅与第一层金属之间扩散和多晶硅，或非晶硅。

对反熔丝进行编程需要额外的电路来提供高编程电压和相对较高的电流5 米A或者更多。这是通过大型晶体管为每个反熔丝提供寻址来完成的。

反熔丝通常是“关闭”设备。只有一小部分需要打开的总数必须编程（大约2%对于一个典型的应用程序）。因此，在其他条件相同的情况下，使用反熔丝进行编程比使用“常开”设备更快。

必须考虑未编程和编程状态的反熔丝可靠性。超过 40 年的时间相关电介质击穿 (TDDB) 可靠性是一个重要的考虑因素。同样重要的是，编程反熔丝的电阻在部件的使用寿命期间保持较低。对 ONO 电介质的分析表明，它们不会随时间增加电阻。此外，未编程的非晶反熔丝的寄生电容明显低于其他编程技术。

计算机代写|数字硬件系统代写Digital hardware systems代考|Logic Cell Architecture

在本节中，我们对当今使用的商用 FPLD 逻辑单元架构进行了调查，包括它们的组合部分和顺序部分。FPLD 逻辑单元在大小和实现能力上都有所不同。一个双晶体管逻辑单元只能实现一个小尺寸的反相器，而查找表逻辑单元可以实现几个输入变量的任何逻辑功能，而且体积要大得多。为了捕获这些差异，我们通常按粒度对逻辑块进行分类。

由于粒度可以通过多种方式定义（如逻辑块可以实现的布尔函数的数量、双输入与门的数量、晶体管的总数等），我们选择将商业块分为两类: 细粒和粗粒。

细粒度逻辑单元类似于 MPLD 基本单元。最细粒度的逻辑单元将与 MPLD 的基本单元相同，并且将由几个可以可编程互连的晶体管组成。

Crosspoint Solutions 的 FPGA 在逻辑单元中使用单个晶体管对。除了晶体管对块外，如图 1.9 所示，交叉点 FPGA 还有第二种类型的逻辑单元，称为 RAM 逻辑块，它针对随机存取存储器的实现进行了调整，但也可用于构建其他逻辑函数。

细粒度 FPGA 架构的第二个示例是 Plessey 的 FPGA。这里的基本单元是一个双输入与非门，如图 1.10 所示。通过连接与非门以实现所需功能，以通常的方式形成逻辑。如果不需要锁存器，则配置存储器设置为使锁存器永久透明。

其他几种商业 FPGA 使用细粒度逻辑单元。使用细粒度逻辑单元的主要优点是可用单元得到充分利用。这是因为更容易高效地使用小型逻辑门，而且此类单元的逻辑综合技术与传统 MPGA（掩模可编程门阵列）和标准单元的逻辑综合技术非常相似。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

计算机代写|数字硬件系统代写Digital hardware systems代考|ECE327

Posted on 2023年2月14日2023年2月14日 by statistics-lab

如果你也在怎样代写数字硬件系统Digital hardware systems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。

我们提供的数字硬件系统Digital hardware systems及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

计算机代写|数字硬件系统代写Digital hardware systems代考|ECE327

计算机代写|数字硬件系统代写Digital hardware systems代考|Static RAM FPGAs

In SRAM FPGAs, static memory cells hold the program that represents the user design. SRAM FPGAs implement logic as lookup tables (LUTs) made from memory cells with function inputs controlling the address lines. Each LUT of $\mathbf{2}^{\mathbf{n}}$ memory cells implements any function of $\mathrm{n}$ inputs. One or more LUTs, combined with flip-flops, form a logic block (LB). LBs are arranged in a two-dimensional array with interconnect segments in channels as shown in Figure 1.4.

Interconnect segments connect to LB pins in the channels and to the other segments in the switch boxes through pass transistors controlled by configuration memory cells. The switch boxes, because of their high complexity, are not full crossbar switches.

An SRAM FPGA program consists of a single long program word. On-chip circuitry loads this word, reading it serially out of an external memory every time power is applied to the chip. The program bits set the values of all configuration memory cells on the chip, thus setting the lookup table values and selecting which segments connect each to the other. SRAM FPGAs are inherently reprogrammable. They can be easily updated providing designers with new capabilities such as reconfigurability.

An antifuse is a two-terminal device that, when exposed to a very high voltage, forms a permanent short circuit (opposite to a fuse) between the nodes on either side. Individual antifuses are small, enabling an antifuse-based architecture to have thousands or millions of antifuses. Antifuse FPGA, as illustrated in Figure 1.5, usually consists of rows of configurable logic elements with interconnect channels between them, much like traditional gate arrays.

The pins on logic blocks (LBs) extend into the channel. An LB is usually a simple gate-level network, which the user programs by connecting its input pins to fixed values or to intereonnect nets. There are antifuses at every wirc-to-pin intersection point in the channel and at all wire-to-wire intersection points where channels intersect.

计算机代写|数字硬件系统代写Digital hardware systems代考|SRAM Programming Technology

SRAM programming technology uses static RAM cells to configure logic and control intersections and paths for signal routing. The configuration is done by controlling pass gates or multiplexers as it is illustrated in Figure 1.6. When a “1” is stored in the SRAM cell in Figure 1.6(a), the pass gate acts as a closed switch and can be used to make a connection between two wire segments. For the multiplexer, the state of the SRAM cells connected to the select lines controls which one of the multiplexers inputs are connected to the output, as shown in Figure 1.6(b). Reprogrammability allows the circuit manufacturer to test all paths in the FPGA by reprogramming it on the tester. The users get well tested parts and $100 \%$ “programming yield” with no design specific test patterns and no “design for testability.” Since on-chip programming is done with memory cells, the programming of the part can be done an unlimited number of times. This allows prototyping to proceed iteratively, re-using the same chip for new design iterations. Reprogrammability has advantages in systems as well. In cases where parts of the logic in a system are not needed simultaneously, they can be implemented in the same reprogrammable FPGA and FPGA logic can be switched between applications.

Besides volatility, a major disadvantage of SRAM programming technology is its large area. At least five transistors are needed to implement an SRAM cell, plus at least one transistor to implement a programmable switch. A typical five-transistor memory cell is illustrated in Figure 1.7. There is no separate RAM area on the chip. The memory cells are distributed among the logic elements they control. Since FPGA memories do not change during normal operation, they are built for stability and density rather than speed. However, SRAM programming technology has two further major advantages; fast-reprogrammability and that it requires only standard integrated circuit process technology.

数字硬件系统代考

计算机代写|数字硬件系统代写Digital hardware systems代考|Static RAM FPGAs

在 SRAM FPGA 中，静态存储单元保存代表用户设计的程序。SRAM FPGA 将逻辑实现为查找表 (LUT)，这些查找表由具有控制地址线的功能输入的存储单元构成。每个 LUT 的2n存储单元实现的任何功能n输入。一个或多个 LUT 与触发器组合在一起，形成一个逻辑块 (LB)。如图 1.4 所示，LB 排列在通道中具有互连段的二维阵列中。

互连段连接到通道中的 LB 引脚，并通过由配置存储单元控制的传输晶体管连接到开关盒中的其他段。由于其高度复杂性，开关盒不是完整的交叉开关。

SRAM FPGA 程序由一个长程序字组成。片上电路加载这个字，每次给芯片加电时从外部存储器中串行读取它。程序位设置芯片上所有配置存储单元的值，从而设置查找表值并选择哪些段相互连接。SRAM FPGA 本质上是可重新编程的。它们可以轻松更新，为设计人员提供新功能，例如可重新配置。

反熔丝是一种双端器件，当暴露于非常高的电压时，会在两侧的节点之间形成永久短路（与熔丝相反）。单个反熔丝很小，使得基于反熔丝的架构可以拥有数千或数百万个反熔丝。反熔丝 FPGA，如图 1.5 所示，通常由多行可配置逻辑元件组成，它们之间有互连通道，很像传统的门阵列。

逻辑块 (LB) 上的引脚延伸到通道中。LB 通常是一个简单的门级网络，用户通过将其输入引脚连接到固定值或互连网络来对其进行编程。在通道中的每个线对引脚交叉点以及通道相交的所有线对线交叉点都有反熔丝。

计算机代写|数字硬件系统代写Digital hardware systems代考|SRAM Programming Technology

SRAM 编程技术使用静态 RAM 单元来配置逻辑并控制信号路由的交叉点和路径。配置是通过控制传输门或多路复用器完成的，如图 1.6 所示。当“1”存储在图 1.6(a) 中的 SRAM 单元中时，传输门充当闭合开关，可用于在两个线段之间建立连接。对于多路复用器，连接到选择线的 SRAM 单元的状态控制多路复用器输入中的哪一个连接到输出，如图 1.6(b) 所示。可重编程性允许电路制造商通过在测试仪上重新编程来测试 FPGA 中的所有路径。用户得到经过良好测试的零件和100%“编程产量”没有设计特定的测试模式，也没有“可测试性设计”。由于片上编程是通过存储单元完成的，因此可以无限次地对部件进行编程。这允许原型设计迭代地进行，为新的设计迭代重复使用相同的芯片。可重编程性在系统中也有优势。在系统中不同时需要部分逻辑的情况下，它们可以在同一个可重新编程的 FPGA 中实现，并且 FPGA 逻辑可以在应用程序之间切换。

除了易失性，SRAM 编程技术的一个主要缺点是它的大面积。至少需要五个晶体管来实现一个 SRAM 单元，再加上至少一个晶体管来实现一个可编程开关。典型的五晶体管存储单元如图 1.7 所示。芯片上没有单独的 RAM 区域。存储单元分布在它们控制的逻辑元件中。由于 FPGA 存储器在正常操作期间不会发生变化，因此它们是为稳定性和密度而不是速度而构建的。然而，SRAM 编程技术还有两个主要优势；快速再编程能力，它只需要标准的集成电路工艺技术。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|CS561

Posted on 2022年7月8日2022年7月8日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

statistics-lab™ 为您的留学生涯保驾护航在代写程序设计algorithm Programming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。我们的专家在代写程序设计algorithm Programming代写方面经验极为丰富，各种代写程序设计algorithm Programming相关的作业也就用不着说。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|CS561

CS代写|程序设计作业代写algorithm Programming代考|Not That Kind of Graph

Your task is to graph the price of a stock over time. In one unit of time, the stock can either Rise, Fall, or stay Constant. The stock’s price will be given to you as a string of R’s, F’s, and C’s. You need to graph it using the characters ‘ $/$ ‘ (slash), ‘ $R$ ‘ (backslash) and ‘_’ (underscore).
Input
The first line of input gives the number of cases, $N$. $N$ test cases follow. Each one contains a string of at least 1 and at most 50 uppercase characters ( $R, F$, or $C$ ).
Output
For each test case, output the line “Case #x:”, where $x$ is the number of the test case. Then print the graph, as shown in the sample output, including the $x$ – and $y$-axes. The $x$-axis should be one character longer than the graph, and there should be one space between the $y$-axis and the start of the graph. There should be no trailing spaces on any line. Do not print unnecessary lines. The $x$-axis should always appear directly below the graph. Finally, print an empty line after each test case.

CS代写|程序设计作业代写algorithm Programming代考|Factorial! You Must be Kidding

Arif has bought a supercomputer from Bongobazar. Bongobazar is a place in Dhaka where secondhand goods are available. So the supercomputer he bought is also secondhand and has some bugs. One of the bugs is that the range of unsigned long integers of this computer for a $\mathrm{C} / \mathrm{C}++$ compiler has changed. Now its new lower limit is 10000 and the upper limit is 6227020800 . Arif writes a program in C/C++ which determines the factorial of an integer. The factorial of an integer is defined recursively as:
Factorial $(0)=1$
Factorial $(n)=n \times$ Factorial $(n-1)$.
Of course, one can manipulate these expressions. For example, it can be written as:
$$
\text { Factorial }(n)=n \times(n-1) \times \text { Factorial }(n-2)
$$
This definition can also be converted to an iterative one.
But Arif knows that his program will not behave correctly in the supercomputer. You are to write a program which will simulate that changed behavior in a normal computer.
Input
The input file contains several lines of input. Each line contains a single integer $n$. No integer has more than 6 digits. Input is terminated by end of file.

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|Not That Kind of Graph

您的任务是绘制股票价格随时间变化的图表。在一个单位时间内，股票可以上涨、下跌或保持不变。股票的价格将以 R、F和C 的字符串形式提供给您。您需要使用字符 ‘ 绘制它/’ (斜线)， ‘ $R$ ‘ (反斜杜) 和’.’ (下划线)。输入
第一行输入给出了案例的数量， $N . N$ 测试用例如下。每个都包含至少 1 个和最多 50 个大写字符的字符串 $(R, F$ 或者 $C)$ 。
输出
对于每个测试用例，输出”Case #x:”行，其中 $x$ 是测试用例的编号。然后打印图形，如示例输出所示，包括 $x$ – 和 $y$ 轴。这 $x$-axis应该比图表长一个字符，并且应该有一个空格。 $y$-轴和图表的开始。任何一行都不应有尾随空格。不要打印不必要的行。这 $x$-axis 应始终显示在图形的正下方。最后，在每个测试用例之后打印一个空行。

CS代写|程序设计作业代写algorithm Programming代考|Factorial! You Must be Kidding

Arif 从 Bongobazar 购买了一台超级计算机。Bongobazar 是达卡的一个出售二手商品的地方。所以他买的超级电脑也是二手的，有一些bug。错误之一是这台计算机的无符号长整数范围 $\mathrm{C} / \mathrm{C}++$ 编译器变了。现在它的新下限是 10000 ，上限是 6227020800 。Arif 用 C/C++ 编写了一个确定整数阶乘的程序。
整数的阶乘递归定义为: $(0)=1$
阶乘 $(n)=n \times$ 阶乘 $(n-1)$.
当然，可以操纵这些表达式。例如，它可以写成:
Factorial $(n)=n \times(n-1) \times$ Factorial $(n-2)$
此定义也可以转换为迭代定义。
但是 Arif 知道他的程序在超级计算机中无法正常运行。您将编写一个程序来模拟普通计算机中的更改行为。输入
输入文件包含几行输入。每行包含一个整数 $n$. 没有整数超过 6 位。输入在文件末尾终止。

CS代写|程序设计作业代写algorithm Programming代考请认准statistics-lab™

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|COMP110

Posted on 2022年7月8日2022年7月8日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|COMP110

CS代写|程序设计作业代写algorithm Programming代考|Tic Tac Toe

Tic Tac Toe is a child’s game played on a 3 by 3 grid. One player, X, starts by placing an $\mathrm{X}$ at an unoccupied grid position. Then the other player, $\mathrm{O}$, places an $\mathrm{O}$ at an unoccupied grid position. Play alternates between $X$ and $O$ until the grid is filled or one player’s symbols occupy an entire line (vertical, horizontal, or diagonal) in the grid.

We will denote the initial empty Tic Tac Toe grid with nine dots. Whenever $\mathrm{X}$ or $\mathrm{O}$ plays, we fill in an $\mathrm{X}$ or an $\mathrm{O}$ in the appropriate position. The example in Figure $1.4$ illustrates each grid configuration from the beginning to the end of game in which $X$ wins.

Your job is to read a grid and to determine whether or not it could possibly be part of a valid Tic Tac Toe game. That is, is there a series of plays that can yield this grid somewhere between the start and end of the game?
Input
The first line of input contains $N$, the number of test cases. $4 N-1$ lines follow, specifying $N$ grid configurations separated by empty lines.

CS代写|程序设计作业代写algorithm Programming代考|Rock, Scissors, Paper

Bart’s sister Lisa has created a new civilization on a two-dimensional grid. At the outset, each grid location may be occupied by one of three life forms: Rocks, Scissors,or Papers. Each day, differing life forms occupying horizontally or vertically adjacent grid locations wage war. In each war, Rocks always defeat Scissors, Scissors always defeat Papers, and Papers always defeat Rocks. At the end of the day, the victor expands its territory to include the loser’s grid position. The loser vacates the position.

Your job is to determine the territory occupied by each life form after $n$ days.
Input
The first line of input contains $t$, the number of test cases. Each test case begins with three integers not greater than $100: r$ and $c$, the number of rows and columns in the grid, and $n$. The grid is represented by the $r$ lines that follow, each with $c$ characters. Each character in the grid is $R, S$, or $P$, indicating that it is occupied by Rocks, Scissors, or Papers respectively.

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|Tic Tac Toe

Tic Tac Toe 是一种儿童游戏，在 3 x 3 网格上进行。一名玩家 X 开始放置一个X在一个未被占用的网格位置。然后另一个玩家，○, 放置一个○在一个未被占用的网格位置。交替播放X和○直到网格被填满或一个玩家的符号占据网格中的整条线（垂直、水平或对角线）。

我们将用九个点表示初始空井字游戏网格。每当X或者○播放，我们填写一个X或一个○在适当的位置。图中的例子1.4说明了从游戏开始到结束的每个网格配置，其中X获胜。

您的工作是阅读网格并确定它是否可能是有效井字游戏的一部分。也就是说，是否有一系列的游戏可以在游戏开始和结束之间的某个地方产生这个网格？
输入输入
的第一行包含ñ，测试用例的数量。4ñ−1接下来的行，指定ñ由空线分隔的网格配置。

CS代写|程序设计作业代写algorithm Programming代考|Rock, Scissors, Paper

巴特的妹妹丽莎在二维网格上创造了一个新文明。一开始，每个网格位置可能被三种生命形式中的一种占据：石头、剪刀或纸。每天，占据水平或垂直相邻网格位置的不同生命形式发动战争。在每场战争中，Rocks 总是击败 Scissors，Scissors 总是击败 Papers，Papers 总是击败 Rocks。在一天结束时，胜利者将其领土扩大到包括失败者的网格位置。失败者腾出位置。

你的工作是确定每个生命形式在之后所占据的领土n天。
输入输入
的第一行包含吨，测试用例的数量。每个测试用例以三个不大于100:r和C，网格中的行数和列数，以及n. 网格由r接下来的行，每行都有C人物。网格中的每个字符都是R,小号，或者磷，表示它分别被 Rocks、Scissors 或 Papers 占据。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|CSE373

Posted on 2022年7月8日2022年7月8日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|CSE373

CS代写|程序设计作业代写algorithm Programming代考|Uniform Generator

Computer simulations often require random numbers. One way to generate pseudorandom numbers is via a function of the form:
$\operatorname{seed}(x+1)=[\operatorname{seed}(x)+S T E P] \% M O D \quad$ where $” \%$ ” is the modulus operator.
Such a function will generate pseudo-random numbers (seed) between 0 and $M O D-1$. One problem with functions of this form is that they will always generate the same pattern over and over. In order to minimize this effect, selecting the STEP and $M O D$ values carefully can result in a uniform distribution of all values between (and including) 0 and $M O D-1 .$

For example, if $S T E P=3$ and $M O D=5$, the function will generate the series of pseudo-random numbers $0,3,1,4,2$ in a repeating cycle. In this example, all of the numbers between and including 0 and $M O D-1$ will be generated every $M O D$ iterations of the function. Note that by the nature of the function to generate the same $\operatorname{seed}(x+1)$ every time, $\operatorname{seed}(x)$ occurs means that if a function will generate all the numbers between 0 and $M O D-1$, it will generate pseudo-random numbers uniformly with every $M O D$ iteration.

If $S T E P=15$ and $M O D=20$, the function generates the series $0,15,10,5$ (or any other repeating series if the initial seed is other than 0 ). This is a poor selection of $S T E P$ and $M O D$ because no initial seed will generate all of the numbers from 0 and $M O D-1$.

Your program will determine whether choices of $S T E P$ and $M O D$ will generate a uniform distribution of pseudo-random numbers.

CS代写|程序设计作业代写algorithm Programming代考|Minesweeper

The game Minesweeper is played on an $n$ by $n$ grid. In this grid are hidden $m$ mines, each at a distinct grid location. The player repeatedly touches grid positions. If a position with a mine is touched, the mine explodes and the player loses. If a position not containing a mine is touched, an integer between 0 and 8 appears, denoting the number of adjacent or diagonally adjacent grid positions that contain a mine. A sequence of moves in a partially played game is illustrated below in Figure 1.3.
Here, $n$ is $8, m$ is 10 , blank squares represent the integer 0 , raised squares represent unplayed positions, and the figures resembling asterisks represent mines. The leftmost image represents the partially played game. From the first image to the second,the player has played two moves, each time choosing a safe grid position. From the second image to the third, the player is not so lucky; he chooses a position with a mine and therefore loses. The player wins if he continues to make safe moves until only $m$ unplayed positions remain; these must necessarily contain the mines.

Your job is to read the information for a partially played game and to print the corresponding board.

The first line of input contains a single positive integer $n \leq 10$. The next $n$ lines rep resent the positions of the mines. Each line represents the contents of a row using $n$ characters: a period indicates an unmined positon, while an asterisk indicates a mined position. The next $n$ lines are each $n$ characters long: touched positions are denoted by an $x$, and untouched positions by a period. The sample input corresponds to the middle section of Figure 1.3.

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|Uniform Generator

计算机模拟通常需要随机数。生成伪随机数的一种方法是通过以下形式的函数：
$\operatorname{seed}(x+1)=[\operatorname{seed}(x)+S T E P] \% M O D \quad$ 在哪里” \%” 是模运算符。
这样的函数将生成 0 到 0 之间的伪随机数（种子） $M O D-1$. 这种形式的函数的一个问题是它们总是会一遍又一遍地生成相同的模式。为了尽量减少这种影响，选择 STEP 和 $M O D$ 仔细考虑值可以导致所有值在 (包括) 0 和 $M O D-1 .$
例如，如果 $S T E P=3$ 和 $M O D=5$ ，该函数将生成一系列伪随机数 $0,3,1,4,2$ 在一个重复的循环中。在这个例子中，所有介于 0 和 $M O D-1$ 将生成每个 $M O D$ 函数的迭代。请注意，由函数的性质来生成相同的 $\operatorname{seed}(x+1)$ 每次， $\operatorname{seed}(x)$ 发生意味着如果一个函数将生成 0 到 $M O D-1$ ，它将均匀地生成伪随机数 $M O D$ 迭代。
如果 $S T E P=15$ 和 $M O D=20$ ，函数生成序列 $0,15,10,5$ (或任何其他重复序列，如果初始种子不是 0 )。这是一个糟糕的选择 $S T E P$ 和 $M O D$ 因为没有初始种子会从 0 和 $M O D-1$.
您的程序将决定是否选择 $S T E P$ 和 $M O D$ 将生成伪随机数的均匀分布。

CS代写|程序设计作业代写algorithm Programming代考|Minesweeper

游戏扫雷是在一个n经过n网格。在这个网格中隐藏米地雷，每个都位于不同的网格位置。玩家反复触碰网格位置。如果触及带有地雷的位置，地雷就会爆炸，玩家输了。如果触摸了不包含地雷的位置，则会出现一个介于 0 和 8 之间的整数，表示包含地雷的相邻或对角相邻网格位置的数量。下图 1.3 说明了部分进行的游戏中的一系列移动。
这里，n是8,米是 10 ，空白方块代表整数 0 ，凸起的方块代表未打过的位置，类似星号的数字代表地雷。最左边的图像代表部分玩过的游戏。从第一张图片到第二张，玩家玩了两步，每次都选择一个安全的格子位置。从第二张到第三张，玩家就没那么幸运了；他选择了一个有地雷的位置，因此输了。如果玩家继续安全移动直到只有米未使用的位置仍然存在；这些必须包含地雷。

您的工作是读取部分玩过的游戏的信息并打印相应的棋盘。

输入的第一行包含一个正整数n≤10. 下一个n线条代表地雷的位置。每行代表一行的内容，使用n字符：句号表示未开采的位置，而星号表示已开采的位置。下一个n线是每个n长字符：触摸位置由X, 和一段未触及的位置。样本输入对应于图 1.3 的中间部分。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|THE ART OF FILE PROCESSING

Posted on 2022年5月26日2022年5月26日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|THE ART OF FILE PROCESSING

CS代写|程序设计作业代写algorithm Programming代考|A file may generally

A file may generally be defined as an organized collection of well-ordered, well-related, and self-contained information held on a stable storage medium. The information in a file is placed in a specific way and read back in a specific way. The information must be kept together as a unit in the same sequence in a well-organized way. The different units of information must bear some relationship with one another for collective consideration. A file is self-contained in that it is complete in all respects. The stable storage medium may be a piece of paper, a magnetic or optical disk, or a magnetic tape or any other medium where the information can be kept for a long time for repeated use without any special aid. Thus, the information bearing the characteristics mentioned above stored in the main memory of a computer will not make a file because the main memory of a computer can hold it only as long as electricity is supplied to the main memory.

Depending on the nature of the information, files can be classified into two basic types: the program file and the data file. A program file is a file that contains a sequential set of instructions in a computer language that can direct a computer in the performance of some specific task. A data file is a collection of records about closely-related or similar entities. However, these types of files should possess all the features stated in the generalized definition above. A record is an ordered collection of the attribute values of an entity. An attribute is any characteristic or feature of an entity that tells something about the entity, where an entity is anything with a physical or conceptual existence. A fact is anything that is true about an entity. To collect facts about an entity, we first decide on some attributes of the entity and procure facts on those attributes. We normally choose a group of entities called an entity set, a collection of items that are considered together for some close relationship. We next select some of the attributes common to all the entities of the set and collect facts on those attributes in a predefined order to form a record for each of them and put all such records together to form the desired file. Let us consider a business enterprise, for instance. The employees of the enterprise form a closely related set of entities. If we consider the attributes EMPLOYEE-CODE-NUMBER, EMPLOYEE-NAME, EMPLOYEE-ADDRESS, DESIGNATION, and SALARY for each of the employees, then the values of these attributes in the mentioned order form a record of an employee of the enterprise and the collection of all the records of the employees of the enterprise forms the desired employee data file (if the records are kept on a stable storage medium).

CS代写|程序设计作业代写algorithm Programming代考|A file is typically considered

A file is typically considered a data file. The task of file processing is the set of activities performed on the records of a file to generate some desired information. Now depending on the organization of the records, the set of operations will vary. We next consider a discussion on file organization. Basically, file organization can be classified into three categories:
(i) Sequential File Organization
(ii) Indexed File Organization
(iii) Hashed/Relative/Random File Organization
Sequential file organization is one in which records are kept in a file, one after another, and processed in the same sequence in which they are written. The term sequential means one after another, and hence the name bears the nature of the organization of the file.

Indexed file organization is one in which sequentially organized records are associated with an index for the purpose of direct access to the records. An index is a special kind of file that contains records consisting of two attribute values, one that is a unique identifying attribute of the records in the sequential file and the other that contains the address of the records in the main file. The identifying attribute is also known as the key attribute or key field. The records in the index are kept in the ascending order of the key field values. When a user wishes to access a record from an indexed file, she initiates a binary search in the index for some key field value and the record found in the search process is then accessed to get the address of the desired record in the main file. Thus, any record in an indexed file can be accessed without reading the preceding or the following records in the file. This saves time and increases the speed of processing, if the number of records to be accessed in

one processing run is less than one-fourth of the total number of records in the file. The disadvantage of such a file organization is that it takes additional disk space for the index, and hence the file organization is more expensive than sequential file organization. The speed of accessing records also varies depending on the organization of the records in the index. For more details about the index file organization, please see any standard textbook on databases or file management systems.

A hashed or relative file organization is also a direct access file organization. In such a file organization, the key field or identifying attribute value is hashed or converted to some location address in the file space relative to the beginning of the file-record positions on the basis of some predefined function. The predefined function is called a hash routine and the method is called hashing. As the hashing is done dynamically during the creation of the file, no extra file space is needed for this purpose, rather, the records can be pointed to directly later by using the same hash function. The only problem with this type of organization is the proper selection of the hash function and its implementation through programming instructions. The programming efficiency of the developers is considered when selecting one of the two direct access file organizations. (The reader is again advised to read a textbook on file/database management systems for further details.) We now study different problems on file processing to illustrate the flowcharts and the algorithms corresponding to their solutions. The following problems involve sequential file organization.

CS代写|程序设计作业代写algorithm Programming代考|A file named EMPFL

Problem 5.3. A file named EMPFL contains the records of the employees of an organization. Each record consists of data: EMP-CODE, EMP-NAME, and BASIC-PAY. The gross salary of an employee is calculated by using the following formula: Gross Salary = Basic Pay + DA + ADA + HRA + MA
where $\quad D A=45 \%$ of Basic Pay
$A D A=18 \%$ of Basic Pay subject to a minimum of $\$ 200$ and a maximum of $\$ 1000$.
HRA $=25 \%$ of Basic Pay subject to a minimum of $\$ 500$ and a maximum of $\$ 5000$.
$M A=10 \%$ of the Basic pay subject to a minimum of $\$ 100$.
The net salary of an employee is calculated by the formula:
Net Salary = Gross Salary $-$ Total Deduction
where Total Deduction $=P F+P T$
where $\quad P F=12 \%$ of Basic Pay
$P T=5 \%$ of Basic Pay subject to a maximum of $\$ 200$.
Develop a flowchart and the algorithm to show how the salary for different employees is calculated to generate the pay slips for the employees of the organization.

Task Analysis. The problem here is to print the pay slip for each of the employees whose records are contained in EMPFL. This can be done by reading the records of the employees one at a time and then calculating the DA, ADA, HRA, and MA values to determine the gross salary and then finding out the amounts of PF and PT for determining the total deduction. The net salary can then be obtained by subtracting the total deduction from the gross salary. The ADA is $18 \%$ of Basic Pay, subject to a minimum of $\$ 200$ and a maximum of $\$ 1,000$. This implies that if the $18 \%$ of the Basic Pay value happens to be less than $\$ 200$, then $\$ 200$ is the ADA amount; if, however, it exceeds $\$ 1,000$, then $\$ 1,000$ is the ADA amount. The conditions for HRA, MA, and PT can be handled in the same way.

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|A file may generally

文件通常可以定义为保存在稳定存储介质上的有序、相关和自包含信息的有组织集合。文件中的信息以特定方式放置并以特定方式读取。信息必须以相同的顺序以组织良好的方式作为一个单元保存在一起。不同的信息单元必须相互之间存在某种关系，以供集体考虑。文件是自包含的，因为它在所有方面都是完整的。稳定的存储介质可以是一张纸、磁盘或光盘、磁带或任何其他可以长时间保存信息以供重复使用而无需任何特殊帮助的介质。因此，

根据信息的性质，文件可以分为两种基本类型：程序文件和数据文件。程序文件是包含计算机语言中的一组顺序指令的文件，这些指令可以指导计算机执行某些特定任务。数据文件是有关密切相关或相似实体的记录集合。但是，这些类型的文件应具备上述通用定义中所述的所有特征。记录是实体属性值的有序集合。属性是实体的任何特征或特征，它讲述了有关实体的某些信息，其中实体是具有物理或概念存在的任何事物。事实是关于实体的任何真实的东西。收集有关实体的事实，我们首先决定实体的某些属性并获取有关这些属性的事实。我们通常选择一组称为实体集的实体，一组被认为具有某种密切关系的项目的集合。接下来，我们选择集合中所有实体共有的一些属性，并以预定义的顺序收集关于这些属性的事实，为每个属性形成一个记录，并将所有这些记录放在一起形成所需的文件。例如，让我们考虑一个商业企业。企业的员工形成了一组密切相关的实体。如果我们考虑每个员工的属性 EMPLOYEE-CODE-NUMBER、EMPLOYEE-NAME、EMPLOYEE-ADDRESS、DESIGNATION 和 SALARY，

CS代写|程序设计作业代写algorithm Programming代考|A file is typically considered

文件通常被认为是数据文件。文件处理的任务是对文件记录执行的一组活动，以生成一些所需的信息。现在，根据记录的组织，操作集会有所不同。我们接下来考虑关于文件组织的讨论。基本上，文件组织可以分为三类：
(i) 顺序文件组织
(ii) 索引文件组织
(iii) 散列/相对/随机文件组织
顺序文件组织是将记录一个接一个地保存在文件中的一种, 并以与写入它们相同的顺序进行处理。顺序一词意味着一个接一个，因此该名称具有文件组织的性质。

索引文件组织是这样一种组织，其中按顺序组织的记录与索引相关联，以便直接访问记录。索引是一种特殊类型的文件，它包含由两个属性值组成的记录，一个是顺序文件中记录的唯一标识属性，另一个包含主文件中记录的地址。标识属性也称为关键属性或关键字段。索引中的记录按关键字段值的升序保存。当用户希望访问索引文件中的记录时，她会在索引中启动对某个关键字段值的二进制搜索，然后访问在搜索过程中找到的记录以获取主文件中所需记录的地址。因此，无需读取文件中前面或后面的记录即可访问索引文件中的任何记录。如果要访问的记录数在

一次处理运行少于文件中记录总数的四分之一。这种文件组织的缺点是索引占用了额外的磁盘空间，因此文件组织比顺序文件组织更昂贵。访问记录的速度也因索引中记录的组织而异。有关索引文件组织的更多详细信息，请参阅有关数据库或文件管理系统的任何标准教科书。

散列或相对文件组织也是直接访问文件组织。在这样的文件组织中，关键字段或标识属性值被散列或转换为文件空间中相对于文件记录位置的开始的某个位置地址，基于一些预定义的函数。预定义的函数称为散列例程，该方法称为散列。由于散列是在文件创建过程中动态完成的，因此不需要额外的文件空间，而是可以在以后使用相同的散列函数直接指向记录。这种组织的唯一问题是正确选择散列函数并通过编程指令实现它。在选择两种直接访问文件组织中的一种时，会考虑开发人员的编程效率。（再次建议读者阅读有关文件/数据库管理系统的教科书以获取更多详细信息。）我们现在研究文件处理的不同问题，以说明流程图和与其解决方案相对应的算法。以下问题涉及顺序文件组织。

CS代写|程序设计作业代写algorithm Programming代考|A file named EMPFL

问题 5.3。名为 EMPFL 的文件包含组织员工的记录。每条记录由以下数据组成：EMP-CODE、EMP-NAME 和 BASIC-PAY。员工的工资总额使用以下公式计算：工资总额 = 基本工资 + DA + ADA + HRA + MA
其中D一种=45%基本工资
一种D一种=18%基本工资的最低限度$200并且最多$1000.
游戏=25%基本工资的最低限度$500并且最多$5000.
米一种=10%基本工资的最低限额$100.
员工的净工资按以下公式计算：
净工资 = 总工资−总扣除
总扣除=磷F+磷吨
在哪里磷F=12%基本工资
磷吨=5%基本工资的最高限额为$200.
开发流程图和算法，以显示如何计算不同员工的工资以生成组织员工的工资单。

任务分析。这里的问题是打印记录包含在 EMPFL 中的每个员工的工资单。这可以通过一次读取一个员工的记录来完成，然后计算 DA、ADA、HRA 和 MA 值以确定总工资，然后找出 PF 和 PT 的金额以确定总扣除额。然后可以通过从总工资中减去总扣除额来获得净工资。ADA是18%基本工资，但须至少$200并且最多$1,000. 这意味着如果18%的基本工资值恰好小于$200，然后$200是 ADA 金额；但是，如果它超过$1,000，然后$1,000是 ADA 金额。HRA、MA 和 PT 的条件可以以相同的方式处理。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|PROBLEMS INVOLVING

Posted on 2022年5月26日2022年5月26日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|PROBLEMS INVOLVING

CS代写|程序设计作业代写algorithm Programming代考|ARRAYS

Think of a road with a row of houses on it. How would you get a unique address for a house on that road? You would take the name of the road and the house number of the lot. An array is similar to a road with a number of houses. The name of the road can be thought of as the name of the array and the number of the house can be thought of as the location number in the array.
Formally speaking, an array is a finite collection of homogeneous data values usually stored in consecutive memory locations with a common name. The term finite implies that the number of data values of an array must be limited by its size. The term homogeneous means “having the same nature or characteristic.” The term usually implies that arrays are almost always implemented by using contiguous locations of the computer’s main memory in a linearly ordered fashion, but not always. The common name assigned to a set of adjacent memory locations to hold the data of a particular type is called the name of the array. The different data values of an array are mentioned by using the name of the array along with a subscript within brackets, such as $\mathrm{A}[1], \mathrm{A}(1)$, and $\mathrm{A}[2]$, or in general, $\mathrm{A}[i]$, where $i$ must be an integer. The value of $i$ is the location. The subscript is also called an index. This is why an array element such as $\mathrm{A}[i]$ is also called an indexed or subscripted variable. The following are some examples of arrays:

The roll numbers of the students of a class stored in a computer’s main memory in linear order

The names of the students of a class stored in the computer’s main memory in linear order
The maximum temperatures of different days of a month in a city stored in the computer’s memory in linear order

All of the data stored together are of the same type, i.e., homogeneous. For example, roll numbers are usually integers, names are usually strings of characters, and temperatures are usually fractional or floating point numbers. Hence, the first example is an array of integers, the second example is an array of strings, and the third example is an array of floating point numbers. Different computer languages use different notations to represent the array elements. However, we will use just one notation. If $\mathrm{A}$ is an array of size $n$, then we will point to an array element by the notation $\mathrm{A}(i)$, where the value of $i$ can vary from 1 to $n$.

Problem 4.1. The goal here is to show you how to construct an array. The following algorithm will clarify the steps:

Decide the size of the array to be formed, say $n$.
Declare an array of size $n$ with some desired name, say A.

CS代写|程序设计作业代写algorithm Programming代考|The inputs are the grades obtained

Task Analysis. The inputs are the grades obtained by students on three tests. To identify the student, the student roll-number and name of each student are given as input. The final score of each student is obtained by determining the greater score of the first two tests and then adding it to that of the third test. The total score represents the percentage score because the total is based on the marks of two tests, each of which carries a maximum grade of 50 . At this stage, we shall have the Roll Number, Name, and Percentage of all the students. The next task is to sort the facts to get information about the students in a descending sequence of percentages. To sort the facts, we take the percentage of the first student and compare it with the percentage of all the other remaining students and interchange the student’s data whenever some student’s percentage is found to be less than the percentage of that of the first student’s percentage. Similarly, we take the percentage of the second student to compare it with the percentage of the third student to interchange the facts, if needed. This type of comparison is continued until we compare the percentage of the last two students.

CS代写|程序设计作业代写algorithm Programming代考|repeat these comparisons

We repeat these comparisons each time, considering one less element than that in the preceding step. It can be observed that after $(N-1)$ steps, the set of numbers will be in the sorted sequence.
The algorithm of the above process is stated below:
Step 1. FOR I = $1 \mathrm{TON}$
Step 2. INPUT TO A(I)
Step 3. END-FOR
Step 4. FOR I = 1 TO N $-1$
Step 5. FOR J=1 TO N – I
Step 6. $\operatorname{IF~} \mathrm{A}(\mathrm{J})>\mathrm{A}(\mathrm{J}+1)$
THEN $\mathrm{T} \leftarrow \mathrm{A}(\mathrm{J})$
$\mathrm{A}(\mathrm{J}) \leftarrow \mathrm{A}(\mathrm{J}+1)$
$\mathrm{A}(\mathrm{J}+1) \leftarrow \mathrm{T}$
END-IF
Step 7. END-FOR-J
Step 8. END-FOR-I
Step 9. FOR I = 1 TO N
Step 10. PRINT A(I)
Step 11. END-FOR-I
Step 12. STOP
Problem 4.10. Draw a flowchart to show how the product of two matrices can be obtained.

Task Analysis. We know that a matrix is a two-dimensional array. The multiplication of two matrices is possible if the number of columns of the first matrix is equal to the number of rows in the second matrix or if the number of rows in the first matrix equals the number of columns of the second matrix. If we consider the row-by-column multiplication of the two matrices, then each element of a row is taken sequentially to multiply with the corresponding column elements, taking one at a time, and the sum of these products is taken as an element of the resulting matrix. This is repeated for all the rows of the first matrix. The reverse process is carried out for the column by row multiplication. To describe the process mathematically, let $\mathrm{A}=\left[a_{i j}\right]$ be an $m \times n$ matrix and $\mathrm{B}=\left[b_{i j}\right]$ be an $n \times p$ matrix. Then the product A.B of these matrices is of the order $m \times p$ say, $\mathrm{C}=\left[c_{i j}\right]$.
where $c_{i j}=a_{i 1} \cdot b_{1 j}+a_{i 2} \cdot b_{2 i}+\ldots \ldots+a_{i n} \cdot b_{n j}$
$$
\Rightarrow \quad c_{i j}=\sum_{k=1}^{n} a_{i k} \cdot b_{k j}
$$

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|ARRAYS

想象一条道路上有一排房子。您如何获得那条路上的房子的唯一地址？你会记下道路的名称和地段的门牌号码。数组类似于带有许多房屋的道路。道路的名称可以认为是数组的名称，房屋的编号可以认为是数组中的位置编号。
正式地说，数组是同质数据值的有限集合，通常存储在具有共同名称的连续内存位置中。术语有限意味着数组的数据值的数量必须受其大小的限制。同质一词的意思是“具有相同的性质或特征”。该术语通常意味着数组几乎总是通过以线性排序方式使用计算机主存储器的连续位置来实现，但并非总是如此。分配给一组相邻内存位置以保存特定类型数据的通用名称称为数组的名称。数组的不同数据值通过使用数组的名称和括号内的下标来提及，例如一种[1],一种(1)，和一种[2]，或者一般来说，一种[一世]，在哪里一世必须是整数。的价值一世是位置。下标也称为索引。这就是为什么数组元素如一种[一世]也称为索引或下标变量。以下是数组的一些示例：

以线性顺序存储在计算机主存储器中的班级学生的卷号

以线性顺序存储在计算机主存储器中的班级学生姓名
以线性顺序存储在计算机内存中的城市一个月内不同日子的最高气温

所有存储在一起的数据都是相同类型的，即同质的。例如，卷号通常是整数，名称通常是字符串，温度通常是小数或浮点数。因此，第一个示例是整数数组，第二个示例是字符串数组，第三个示例是浮点数数组。不同的计算机语言使用不同的符号来表示数组元素。但是，我们将只使用一种表示法。如果一种是一个大小数组n，然后我们将通过符号指向一个数组元素一种(一世)，其中的值一世可以从 1 到n.

问题 4.1。这里的目标是向您展示如何构造一个数组。以下算法将阐明这些步骤：

决定要形成的数组的大小，比如说n.
声明一个大小数组n有一些想要的名字，比如 A。

CS代写|程序设计作业代写algorithm Programming代考|The inputs are the grades obtained

任务分析。输入是学生在三个测试中获得的成绩。为了识别学生，每个学生的学生卷号和姓名作为输入给出。每个学生的最终分数是通过确定前两个测试的较大分数然后将其添加到第三个测试的分数而获得的。总分代表百分比分数，因为总分基于两个测试的分数，每个测试的最高等级为 50 。在这个阶段，我们将获得所有学生的卷号、姓名和百分比。下一个任务是对事实进行排序，以按百分比降序获取有关学生的信息。为了整理事实，我们取第一个学生的百分比并将其与所有其他剩余学生的百分比进行比较，并在发现某个学生的百分比小于第一个学生的百分比时交换学生的数据。同样，如果需要，我们将第二个学生的百分比与第三个学生交换事实的百分比进行比较。这种比较一直持续到我们比较最后两个学生的百分比。

CS代写|程序设计作业代写algorithm Programming代考|repeat these comparisons

我们每次都重复这些比较，考虑比上一步中的元素少一个。可以观察到，之后(ñ−1)步骤，这组数字将按排序顺序排列。
上述过程的算法如下：
Step 1. FOR I =1吨这ñ
步骤 2. 输入 A(I)
步骤 3. END-FOR
步骤 4. FOR I = 1 TO N−1
第 5 步。对于 J=1 TO N – I
第 6 步。一世F ⁡一种(Ĵ)>一种(Ĵ+1)
然后吨←一种(Ĵ)
一种(Ĵ)←一种(Ĵ+1)
一种(Ĵ+1)←吨
END-IF
步骤 7. END-FOR-J
步骤 8. END-FOR-I
步骤 9. FOR I = 1 TO N
步骤 10. 打印 A(I)
步骤 11. END-FOR-I
步骤 12. 停止
问题 4.10 . 画一个流程图来说明如何得到两个矩阵的乘积。

任务分析。我们知道矩阵是一个二维数组。如果第一个矩阵的列数等于第二个矩阵的行数，或者如果第一个矩阵的行数等于第二个矩阵的列数，则两个矩阵相乘是可能的。如果我们考虑两个矩阵的逐列相乘，那么每一行的每个元素都依次与对应的列元素相乘，一次取一个，将这些乘积之和作为一个元素结果矩阵。这对第一个矩阵的所有行重复。对逐行乘法执行相反的过程。为了用数学方法描述这个过程，让一种=[一种一世j]豆米×n矩阵和乙=[b一世j]豆n×p矩阵。那么这些矩阵的乘积 AB 是有序的米×p说，C=[C一世j].
在哪里C一世j=一种一世1⋅b1j+一种一世2⋅b2一世+……+一种一世n⋅bnj

⇒C一世j=∑ķ=1n一种一世ķ⋅bķj

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

时间序列分析代写

回归分析代写

MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|Construct a flowchart to show

Posted on 2022年5月26日2022年5月26日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|Construct a flowchart to show

Problem 3.23. Construct a flowchart to show how you can decide if a given number is prime or not.

Task Analysis. We know that a number can be called a prime number if, and only if, it has no divisor or factor except itself and unity, i.e., 1. In order to declare that a number is a prime number, we need to prove that the number is not divisible by any number starting from 2 to the half of the given number because we have already seen that if a number has some divisor at all, it must lie within the half of the number. A better, more efficient strategy is to limit the checking within the integer part of the square root of the number. For example, to check if the number 97 is a prime number, we need check whether there exists some divisor of 97 within 2 to 48 (both inclusive). This checking can be done from 2 to 9 , because 9 is the integer part of the square root of 97 . The number of checking is decreased to a large extent. The divisors can be generated automatically by changing the value of a variable location. Assuming that the procedure for determining the square root of a number is available, we can draw the flowchart for the task.

The following algorithm shows the steps leading to the solution for Problem 3.23:
Step 1. INPUT TO N
[ACCEPT THE NUMBER WHOSE SQUARE ROOT IS TO BE FOUND]
Step 2. COMPUTE SR $\leftarrow$ SQUARE ROOT OF (N)
Step 3. [INITIALIZE] I $\leftarrow 2, \mathrm{FLAG} \leftarrow 0$
[FLAG contains the divisibility status of the number]
Step 4. WHILE I $<=S R$ DO
(i) COMPUTE $\mathrm{R} \leftarrow$ REMAINDER OF (N/I)
(ii) IF $\mathrm{R}=0$
THEN FLAG $\leftarrow 1$
EXIT
END-IF
(iii) COMPUTE I $\leftarrow$ I $+1$
(Increment I to obtain the next divisor]
Step 5. IF FLAG =0
THEN PRINT “It is a prime number.”
ELSE
PRINT”It is not a prime number.”
END-IF
Step 6. STOP

CS代写|程序设计作业代写algorithm Programming代考|The algorithm below shows the solution of Problem

The algorithm below shows the solution of Problem $3.24$.
Step 1. INPUT TO N
[Establish N, the number of FIBONACCI NUMBERS to be generated]
Step 2. [INITIALIZE VARIABLES WITH THE FIRST TWO FIBONACCI NUMBERS]
$\mathrm{T} 1 \leftarrow 0, \quad \mathrm{~T} 2 \leftarrow 1$
Step 3. [INITIALIZE THE COUNTER VARIABLE]
COUNT $\leftarrow 0$
Step 4. WHILE COUNT $<=\mathrm{N}$
(i) COMPUTE $\mathrm{T} \leftarrow \mathrm{T} 1+\mathrm{T} 2$
(ii) PRINT Tl
(iii) COMPUTE COUNT $\leftarrow$ COUNT $+1$
(iv) $\mathrm{T} 1 \leftarrow \mathrm{T} 2$
(v) $\mathrm{T} 2 \leftarrow \mathrm{T}$
Step 5. STOP
Problem 3.25. Construct a flowchart to show if a given year is leap year.
Task Analysis. A given year is said to be a leap year if it is a non-century year (i.e., not like 1900,1800 , or 1600 ) and it is divisible by 4 . In case it is a century year, then it must be divisible by 400 to be a leap year. To determine whether a given year is a leap year, we determine whether the year is divisible by 4 but not by 100 or if it is divisible by 400 . The divisibility is tested again in the way as we have seen earlier, i.e., by checking whether the remainder in the division process is zero or not.
Step 1. $\quad \mathrm{Y} \leftarrow 1$
Step 2. REPEAT STEPS 2 TO 8 UNTIL $\mathrm{Y}=0$
Step 3. INPUT TO Y
[ACCEPT YEAR TO BE TESTED AND STORE IT IN Y]
Step 4. IF $Y=0$
THEN EXIT
END-IF
Step 5. COMPUTE R $1 \leftarrow$ REMAINDER OF (Y/400)
Step 6. $\quad$ IF Rl $=0$
THEN PRINT “THE GIVEN YEAR IS A LEAP YEAR” END-IF

CS代写|程序设计作业代写algorithm Programming代考|The square root of a number

Task Analysis. The square root of a number can be obtained by using the Newton Raphson Method. In this method, the square root of any positive number is initially set to 1 . Then the absolute value of the difference between

the square of the assumed square root and the given number is obtained. This value is then compared with some predefined small positive number. This small positive number is set in such a way that an error of magnitude less than that is made acceptable. If the difference is less than the small positive number, the assumed square root is used as the desired square root. For perfect squares, this difference becomes zero; for others, this difference is usually found to be of magnitude less than $.01, .001$, or .0001, depending upon the precision required. If the difference is greater than or equal to the small positive number like $.001$ or $.0001$, then the assumed value is increased to have a better guess by using the formula
$$
\left(\text { Guessed Value }+\frac{\text { Number }}{\text { Guessed Value }}\right) / 2
$$
The procedure is repeated until we get a guessed value satisfying the condition specified. Algorithmically, we can express the procedure as shown below.
Let X be the number whose square root is to be obtained.

Set Guess to $1 .$
If $\mid$ GUESS*GUESS-X $\mid<$ Epsilon
Then go to step 5
(Epsilon is a predefined small positive number)
Set Guess to $\left(\right.$ Guess $\left.+\frac{\mathrm{X}}{\text { Guess }}\right) / 2$
Go to Step 2
Guess is the square root of X.
The flowchart corresponding to Problem $3.26$ is shown in next page. The algorithm for the solution of Problem $3.26$ is given below:
Step 1. INPUT TO X
Step 2. [INITIALIZE] GUESS $\leftarrow 1$, EPSILON $\leftarrow .001$
Step 3. WHILE absolute value of $(\mathrm{GUESS} * \mathrm{GUESS}-\mathrm{X})<=$ EPSILON DO
$\mathrm{COMPUTE} \mathrm{GUESS} \leftarrow\left(\mathrm{GUESS}+\frac{\mathrm{X}}{\mathrm{GUESS}}\right) / 2$
Step 4. PRINT “THE SQUARE ROOT IS”, GUESS
Step 5. STOP

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|Construct a flowchart to show

问题 3.23。构建一个流程图来展示如何确定给定数字是否为素数。

任务分析。我们知道，一个数可以称为素数，当且仅当它除了自身和一，没有除数或因数，即 1。为了声明一个数是素数，我们需要证明number 不能被从 2 到给定数字一半的任何数字整除，因为我们已经看到，如果一个数字有任何除数，它必须在该数字的一半之内。一个更好、更有效的策略是将检查限制在数字平方根的整数部分内。例如，要检查数字 97 是否为质数，我们需要检查 2 到 48（均包括）内是否存在 97 的除数。这种检查可以从 2 到 9 进行，因为 9 是 97 的平方根的整数部分。检查次数大大减少。除数可以通过更改变量位置的值自动生成。假设确定一个数的平方根的过程是可用的，我们可以画出该任务的流程图。

以下算法显示了求解问题 3.23 的步骤：
步骤 1. 输入 N
[接受要求平方根的数]
步骤 2. 计算 SR←(N) 的
平方根步骤 3. [初始化] I←2,F大号一种G←0
[FLAG 包含数字的整除状态]
Step 4. WHILE I<=小号R做
(i) 计算R←(N/I)
(ii) IF的剩余部分R=0
然后标志←1
退出
END-IF
(iii) 计算 I←一世+1
（增加 I 以获得下一个除数）
Step 5. IF FLAG =0
THEN PRINT “It is a prime number.”
ELSE
PRINT“It is not a prime number.”
END-IF
Step 6. STOP

CS代写|程序设计作业代写algorithm Programming代考|The algorithm below shows the solution of Problem

下面的算法显示了问题的解决方案3.24.
步骤 1. 输入 N
[建立 N，要生成的斐波那契数]
步骤 2. [使用前两个斐波那契数初始化变量]
吨1←0, 吨2←1
步骤 3. [初始化计数器变量]
计数←0
第 4 步。同时计数<=ñ
(一) 计算吨←吨1+吨2
(ii) 打印 Tl
(iii) 计算计数←数数+1
(四)吨1←吨2
（在）吨2←吨
步骤 5. 停止
问题 3.25。构建流程图以显示给定年份是否为闰年。
任务分析。如果给定年份是非世纪年份（即，不像 1900,1800 或 1600 ）并且可以被 4 整除，则称该年份为闰年。如果是世纪年，那么它必须能被 400 整除才能成为闰年。为了确定给定年份是否为闰年，我们确定该年份是否能被 4 整除但不能被 100 整除，或者它是否能被 400 整除。可除性再次以我们之前看到的方式进行测试，即通过检查除法过程中的余数是否为零。
第1步。是←1
第 2 步。重复第 2 到 8 步，直到是=0
步骤 3. 输入 Y
[接受要测试的年份并将其存储在 Y]
步骤 4. 如果是=0
然后退出
END-IF
步骤 5. 计算 R1←(Y/400)
步骤 6 的剩余部分。IF R1=0
然后打印“给定的年份是闰年”END-IF

CS代写|程序设计作业代写algorithm Programming代考|The square root of a number

任务分析。一个数的平方根可以通过使用牛顿拉夫森方法获得。在此方法中，任何正数的平方根最初都设置为 1 。那么两者之差的绝对值

获得假定平方根和给定数字的平方。然后将该值与一些预定义的小正数进行比较。这个小的正数以这样的方式设置，使得小于该大小的误差是可以接受的。如果差值小于小正数，则使用假定的平方根作为所需的平方根。对于完美的正方形，这个差变为零；对于其他人，通常发现这种差异的幅度小于.01,.001，或 .0001，取决于所需的精度。如果差值大于或等于小的正数，例如.001或者.0001，然后通过使用公式增加假设值以获得更好的猜测

( 猜测值 + 数字猜测值 )/2
重复该过程，直到我们得到满足指定条件的猜测值。在算法上，我们可以表达如下所示的过程。
令 X 为要获得其平方根的数。

将猜测设置为1.
如果∣GUESS*GUESS-X∣<Epsilon
然后转到第 5 步
（Epsilon 是一个预定义的小正数）
将猜测设置为(猜测+X 猜测 )/2
转到第 2 步
Guess 是 X 的平方根
。Problem 对应的流程图3.26显示在下一页。解决问题的算法3.26如下所示：
步骤 1. 输入到 X
步骤 2. [初始化] GUESS←1, 爱普生←.001
步骤 3. WHILE 的绝对值(G在和小号小号∗G在和小号小号−X)<=爱普生做
C这米磷在吨和G在和小号小号←(G在和小号小号+XG在和小号小号)/2
第 4 步。打印“平方根是”，猜猜
第 5 步。停止

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金融工程代写

非参数统计代写

非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。

广义线性模型代考

广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。

有限元方法代写

随机分析代写

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MATLAB代写

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写

CS代写|程序设计作业代写algorithm Programming代考|Natural numbers

Posted on 2022年5月26日2022年5月26日 by statistics-lab

如果你也在怎样代写程序设计algorithm Programming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。

程序设计是用于解决一个问题的程序或公式。它是基于进行一连串指定的行动，这些行动描述了如何做某事，而你的计算机每次都会以这种方式做。

我们提供的程序设计algorithm Programming及其相关学科的代写，服务范围广, 其中包括但不限于:

Statistical Inference 统计推断
Statistical Computing 统计计算
Advanced Probability Theory 高等概率论
Advanced Mathematical Statistics 高等数理统计学
(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

CS代写|程序设计作业代写algorithm Programming代考|Natural numbers

Task Analysis. Natural numbers are those numbers that are obtained through sequential counting. The starting number here for the summation process is 1 , the next number is 2 and so on, until we reach $n$-the number of natural numbers to be summed. The numbers to be added are known as inputs and can be generated by instructing the computer. We assign the value 1 to a variable to simulate the first natural number. We then add the value of the variable to an accumulator. The accumulator must then contain some initial value to make the summation process semantically correct, i.e., meaningful. This initial value must be 0 in this case because we are adding the first number. We can then increase the value of the variable containing the first natural number by 1 . This next number, which is 2 in this case, can then be added to the current value of the accumulator to obtain the sum of first two natural numbers. In this way, we can continue the generation and summation process until we add up all the natural numbers, including $N$, for some given value of $\mathrm{N}$. But we must also keep a count of the numbers that are being added; otherwise we will not be able to decide whether we have added the desired $N$ numbers or not. A variable is used here as a counter. This counter must be initialized to zero first, from which we can increment its value each time by 1 when we add some number to the value of the accumulator.

The algorithm corresponding to Problem $3.6$ is shown below:
Step 1. INPUT “ENTER NUMBER OF TERMS TO ADD” TO N
Step 2. SUM $\leftarrow 0$ [INITIALIZATION]
Step 3. I $\leftarrow 1$ [INITIALIZATION]
Step 4. REPEAT STEPS 5 THROUGH 6 WHILE I <= N.
Step 5. COMPUTE SUM $\leftarrow \mathrm{SUM}+\mathrm{I}$
Step 6. COMPUTE I $\leftarrow \mathrm{I}+1$
Step 7. PRINT “THE SUM IS”, SUM
Step 8. STOP
Problem 3.7. Draw a flowchart to show how to obtain the sum of the first 30 natural numbers.

Task Analysis. This problem is similar to Problem 3.6. The only difference is that the number of natural numbers to be added up is given as a constant (30). We do not need input from the user.

CS代写|程序设计作业代写algorithm Programming代考|We require the product of the first

Task Analysis. We require the product of the first 10 natural numbers. The natural numbers are defined in the task analysis of Problem $3.6$, so the natural numbers can be generated similarly. To hold the product, we require a location that is initialized with 1 so that we can specify how to obtain the new product by multiplying the current product by the natural number currently in use. This is because only the initial value 1 will keep the content of the location for the product unchanged when the value of the product location is multiplied by $1 .$
The algorithm showing solution to Problem $3.8$ is as follows:
Step 1. PRODUCT $\leftarrow 1, \mathrm{NUM} \leftarrow 1, \mathrm{CNT} \leftarrow 0$
(Initialize the variables required)
Step 2. REPEAT STEPS 3 THROUGH 5 WHILE CNT $<=10$
Step 3. COMPUTE PRODUCT $\leftarrow$ PRODUCT*NUM
Step 4. COMPUTE CNT $\leftarrow \mathrm{CNT}+1$
(Increment the Counter)
Step 5. COMPUTE NUM $\leftarrow \mathrm{NUM}+1$ (The next number is generated)
Step 6. PRINT “THE PRODUCT IS”, PRODUCT
Step 7. STOP
Problem 3.9. Draw a flowchart to find the sum of first 15 even natural numbers.

Task Analysis. We know that the first natural even number is 2 and the next natural even number, i.e., the second even number, can be obtained by adding 2 to the first natural number. The successive natural even numbers can be obtained by adding 2 to the preceding natural even number. These even numbers can be accumulated in a location by adding the generated even number each time to the accumulator, which contains zero.

A count of the numbers added will enable us to check whether first 15 even natural numbers have been added up or not. No input is required from the user during the time of execution.
The algorithm showing the solution of Problem 3.9. is given below:
Step 1. [Initialize the accumulator, counter and variable]
$\mathrm{SUMM} \leftarrow 0, \mathrm{CNT} \leftarrow 0, \mathrm{NUM} \leftarrow 2$
Step 2. REPEAT STEPS 3 THROUGH 5 WHILE CNT $<15$
Step 3. COMPUTE SUMM $\leftarrow \mathrm{SUMM}+\mathrm{NUM}$
Step 4. COMPUTE CNT $\leftarrow \mathrm{CNT}+1$
Step 5. COMPUTE NUM $\leftarrow \mathrm{NUM}+2$
Step 6. PRINT “THE DESIRED SUM IS”, SUMM
Step 7. STOP

CS代写|程序设计作业代写algorithm Programming代考|The solution of Problem

Problem 3.15. Draw a flowchart to show how to find all even natural numbers that are divisible by 7 in a given range.

Task Analysis. We require two numbers that can serve as boundary values between all the desired numbers to be generated. If a number within the given range is divisible by 7 , then it is printed. As the range may include many numbers, each of the numbers need not be accepted as input from the terminal because it will slow down the whole process. We can generate natural numbers one by one based on the lower range given, and then we test the divisibility by 7 . A number is said to be divisible by 7 if it leaves no remainder when divided by 7 . The input is the numbers forming the lower and the upper ranges between which we test all the numbers, including the numbers forming the ranges. A loop is required to perform the same task of divisibility checking with a newly generated number.

Problem 3.16. Construct a flowchart to find the sum of the squares of the first 9 natural numbers that are divisible by $3 .$

Task Analysis. The problem requires the natural numbers divisible by 3 to obtain their square values and then to accumulate 9 such consecutive square values as the sum of the values.

Our procedure to obtain the sum should encompass generating natural numbers one by one, testing each for divisibility by 3 . If one is found to be divisible, we need to obtain the square of the number to determine the desired sum.

程序设计代写

CS代写|程序设计作业代写algorithm Programming代考|Natural numbers

任务分析。自然数是通过顺序计数获得的那些数字。求和过程的起始数字是 1 ，下一个数字是 2 等等，直到我们到达n- 要求和的自然数的个数。要添加的数字称为输入，可以通过指示计算机生成。我们将值 1 分配给一个变量来模拟第一个自然数。然后我们将变量的值添加到累加器中。然后累加器必须包含一些初始值以使求和过程在语义上正确，即有意义。在这种情况下，此初始值必须为 0，因为我们要添加第一个数字。然后我们可以将包含第一个自然数的变量的值增加 1 。然后可以将下一个数字（在本例中为 2）与累加器的当前值相加，以获得前两个自然数的总和。这样，我们可以继续生成和求和过程，直到我们将所有自然数相加，包括ñ, 对于某个给定的值ñ. 但是我们还必须对正在添加的数字进行计数；否则我们将无法决定是否添加了所需的ñ数字与否。变量在这里用作计数器。该计数器必须首先初始化为零，当我们将一些数字添加到累加器的值时，我们可以每次将其值增加 1。

问题对应的算法3.6如下所示：
Step 1. 输入“ENTER NUMBER OF TERMS TO ADD”到 N
Step 2. SUM←0[初始化]
步骤 3. I←1[初始化]
第 4 步。在 I <= N 时重复第 5 步到第 6
步。第 5 步。计算总和←小号在米+一世
第 6 步。计算 I←一世+1
第 7 步。打印“总和是”，求和
第 8 步。停止
问题 3.7。画一个流程图来说明如何获得前 30 个自然数之和。

任务分析。这个问题与问题 3.6 类似。唯一的区别是要相加的自然数的数量以常数 (30) 的形式给出。我们不需要用户的输入。

CS代写|程序设计作业代写algorithm Programming代考|We require the product of the first

任务分析。我们需要前 10 个自然数的乘积。自然数在问题的任务分析中定义3.6，所以自然数可以类似地生成。为了保存产品，我们需要一个初始化为 1 的位置，以便我们可以通过将当前产品乘以当前使用的自然数来指定如何获得新产品。这是因为当产品位置的值乘以时，只有初始值 1 会保持产品位置的内容不变1.
显示问题解决方案的算法3.8如下：
Step 1. PRODUCT←1,ñ在米←1,Cñ吨←0
（初始化所需的变量）
步骤 2. 重复步骤 3 到 5 WHILE CNT<=10
步骤 3. 计算乘积←PRODUCT*NUM
步骤 4. 计算 CNT←Cñ吨+1
（增加计数器）
步骤 5. COMPUTE NUM←ñ在米+1（生成下一个数字）
第 6 步。打印“产品是”，产品
第 7 步。停止
问题 3.9。画一个流程图，求前 15 个偶数自然数之和。

任务分析。我们知道，第一个自然偶数是2，第一个自然数加2可以得到下一个自然偶数，即第二个偶数。连续的自然偶数可以通过在前面的自然偶数上加2来获得。这些偶数可以通过每次将生成的偶数添加到包含零的累加器中来累积在一个位置。

添加数字的计数将使我们能够检查前 15 个偶数自然数是否已相加。在执行期间不需要用户输入。
显示问题 3.9 解决方案的算法。如下：
Step 1. [初始化累加器、计数器和变量]
小号在米米←0,Cñ吨←0,ñ在米←2
第 2 步。在 CNT 时重复第 3 步到第 5 步<15
步骤 3. 计算总和←小号在米米+ñ在米
步骤 4. 计算 CNT←Cñ吨+1
步骤 5. 计算 NUM←ñ在米+2
步骤 6. 打印“所需的总和是”，求和
步骤 7. 停止

CS代写|程序设计作业代写algorithm Programming代考|The solution of Problem

问题 3.15。画一个流程图来说明如何找出给定范围内能被 7 整除的所有偶数自然数。

任务分析。我们需要两个数字，它们可以作为要生成的所有所需数字之间的边界值。如果给定范围内的数字可以被 7 整除，则打印该数字。由于该范围可能包含许多数字，因此不需要将每个数字都作为来自终端的输入接受，因为它会减慢整个过程。我们可以根据给定的下限一一生成自然数，然后我们测试被 7 整除的能力。如果一个数除以 7 时没有余数，则称该数可被 7 整除。输入是形成上下范围的数字，我们测试所有数字之间的范围，包括形成范围的数字。需要一个循环来执行与新生成的数字的可除性检查相同的任务。

问题 3.16。构造一个流程图，求前 9 个可被 5 整除的自然数的平方和3.

任务分析。该问题要求能被3整除的自然数得到它们的平方值，然后将9个这样的连续平方值累加为这些值的总和。

我们获得总和的过程应该包括一个一个地生成自然数，测试每个自然数的可被 3 整除性。如果发现一个是可整除的，我们需要获得该数字的平方以确定所需的总和。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

R语言代写	问卷设计与分析代写
PYTHON代写	回归分析与线性模型代写
MATLAB代写	方差分析与试验设计代写
STATA代写	机器学习/统计学习代写
SPSS代写	计量经济学代写
EVIEWS代写	时间序列分析代写
EXCEL代写	深度学习代写
SQL代写	各种数据建模与可视化代写