Consider the function f(x) = −x 3 + 60x 2 − 900x − 1. Now imagine that the domain of x is the set of real numbers R. Perform 4 iterations of gradient descent algorithm starting from x = 0, with the following settings: step size η = 0.1 step size η = 1

时间: 2024-04-23 16:22:31 浏览: 16
Sure, I can help you with that! For this problem, we will need to use the gradient descent algorithm to find the minimum of the function f(x) = −x^3 + 60x^2 − 900x − 1. The gradient descent algorithm works by iteratively updating the value of x based on the negative gradient of the function at that point. The update equation is as follows: x := x - η * df/dx where df/dx is the derivative of the function with respect to x and η is the step size. Let's start by performing 4 iterations of gradient descent starting from x = 0 and with step size η = 0.1. Iteration 1: x := 0 - 0.1 * (-3*0^2 + 120*0 - 900) = 90 f(x) = -90^3 + 60*90^2 - 900*90 - 1 = -729001 Iteration 2: x := 90 - 0.1 * (-3*90^2 + 120*90 - 900) = 54 f(x) = -54^3 + 60*54^2 - 900*54 - 1 = -262189 Iteration 3: x := 54 - 0.1 * (-3*54^2 + 120*54 - 900) = 44.4 f(x) = -44.4^3 + 60*44.4^2 - 900*44.4 - 1 = -207764 Iteration 4: x := 44.4 - 0.1 * (-3*44.4^2 + 120*44.4 - 900) = 40.32 f(x) = -40.32^3 + 60*40.32^2 - 900*40.32 - 1 = -188545 Now let's perform 4 iterations of gradient descent starting from x = 0 and with step size η = 1. Iteration 1: x := 0 - 1 * (-3*0^2 + 120*0 - 900) = 900 f(x) = -900^3 + 60*900^2 - 900*900 - 1 = -729000001 Iteration 2: x := 900 - 1 * (-3*900^2 + 120*900 - 900) = -269100 f(x) = -(-269100)^3 + 60*(-269100)^2 - 900*(-269100) - 1 = -2.9628e+24 Iteration 3: x := -269100 - 1 * (-3*(-269100)^2 + 120*(-269100) - 900) = 8.4441e+10 f(x) = -(8.4441e+10)^3 + 60*(8.4441e+10)^2 - 900*(8.4441e+10) - 1 = -6.5828e+32 Iteration 4: x := 8.4441e+10 - 1 * (-3*(8.4441e+10)^2 + 120*(8.4441e+10) - 900) = -2.6349e+22 f(x) = -(-2.6349e+22)^3 + 60*(-2.6349e+22)^2 - 900*(-2.6349e+22) - 1 = -1.2977e+69 As we can see from the results, using a larger step size (η = 1) can cause the algorithm to diverge and not find the minimum of the function. In this case, we need to choose a smaller step size (η = 0.1) in order to converge to the minimum.

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3)A digital clock consists of a screen to display the time and a dial for setting in turn the year, month, day, hour and minute. Twisting the dial to the left reduces by one the value being changed but twisting it to the right increases it by one. Pushing the dial alters which value is being adjusted. At first, it is the year but after the dial is pushed once, it is the month, then after the dial is pushed again, it is the day and so on. Imagine the clock is represented by a class with attributes year, month, day etc. The following is what the code for a method rotateDialLeft() might look like. public void rotateDialLeft() { if (mode == YEAR_MODE) { year--; } else if (mode == MONTH_MODE) { month--; } else if (mode == DAY_MODE) { day--; } else if (mode == HOUR_MODE) { hour--; } else if (mode == MINUTE_MODE) { minute--; } } The code for rotateDialRight() is similar. Apply the Open-Closed Principle to explain why the above code is unsatisfactory from the design viewpoint, considering the possibility of future change to the code, giving an example of such a change. 5)Give the code required for the classes introduced in question 3), focusing on the code for a method selectState() which changes the value that is being adjusted from years to months. Make it clear in which classes the code is to be found. Assume the existence of other methods that are needed such as getMonthSetUpState(). 8)Suppose that in a multiplayer role-playing game, a class Client has a dependency to an interface Fighter with public methods attack(), defend() and escape(). The game designer now wishes for Client to use a class Wizard with three different but equivalent public methods castDestructionSpell(), shield() and portal(). Explain how it is possible to do this using an appropriate design pattern.

Another example is the SRIOV_NET_VF resource class, which is provided by SRIOV-enabled network interface cards. In the case of multiple SRIOV-enabled NICs on a compute host, different qualitative traits may be tagged to each NIC. For example, the NIC called enp2s0 might have a trait “CUSTOM_PHYSNET_PUBLIC” indicating that the NIC is attached to a physical network called “public”. The NIC enp2s1 might have a trait “CUSTOM_PHYSNET_INTRANET” that indicates the NIC is attached to the physical network called “Intranet”. We need a way of representing that these NICs each provide SRIOV_NET_VF resources but those virtual functions are associated with different physical networks. In the resource providers data modeling, the entity which is associated with qualitative traits is the resource provider object. Therefore, we require a way of representing that the SRIOV-enabled NICs are themselves resource providers with inventories of SRIOV_NET_VF resources. Those resource providers are contained on a compute host which is a resource provider that has inventory records for other types of resources such as VCPU, MEMORY_MB or DISK_GB. This spec proposes that nested resource providers be created to allow for distinguishing details of complex components of some resource providers. During review the question came up about “rolling up” amounts of these nested providers to the root level. Imagine this scenario: I have a NIC with two PFs, each of which has only 1 VF available, and I get a request for 2 VFs without any traits to distinguish them. Since there is no single resource provider that can satisfy this request, it will not select this root provider, even though the root provider “owns” 2 VFs. This spec does not propose any sort of “rolling up” of inventory, but this may be something to consider in the future. If it is an idea that has support, another BP/spec can be created then to add this behavior.

clc; clear all; [filename,pathname]=uigetfile({'*.jpg;*.tif;*.png;*gif','all imagine files';'*.*','all files'},'select your photo'); path=[pathname,filename]; image=imread(path); % axes(handles.photo); imshow(image);%显示图片 %image processing I=rgb2gray(image); I=rangefilt(I);%滤波 background = imopen(I,strel('disk',11));%使用形态学开来估计背景 I2 = I-background;%从原始图像中减去背景图像 I3 = imadjust(I2);%增强对比度 bw = imbinarize(I3);%阈 值图像 bw = bwareaopen(bw,160);%降噪150,160 bw=edge(bw,'canny'); %边缘检测 %bw=1-bw; % axes(handles.a1); imshow(bw); %se=strel('disk',13);%15 se=strel('square',15);%15 bw1=imclose(bw,se);%闭 bw2=imdilate(bw1,se);%膨胀 bw2=imerode(bw2,se);%腐蚀 bw3=imfill(bw2,'holes'); % axes(handles.a2); imshow(bw3); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %circle detection rmin = 20; rmax = 2500; radiusRange=[rmin rmax]; [center, rad] = imfindcircles(bw3,radiusRange,'EdgeThreshold',0.13);%检测灵敏度(边缘渐变阈值)0.3 display(center); display(rad); % axes(handles.a3); imshow(bw3); viscircles(center, rad,'Color','b'); %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %initialize the number of coins one=0; half=0; little=0; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %coin recognition [m,n]=size(rad); num=m; i=1; j=num; min=rad(i); max=rad(j); while i<=j if rad(i)<rad(j) if rad(i)<min min=rad(i); else if rad(j)<max max=rad(j); end end else if rad(j)<min min=rad(j); else if rad(i)<max max=rad(i); end end end i=i+1; j=j-1; end sum=0; for i=1:num sum=rad(i)+sum; end % ave=(sum-(min+max))/(num-2); ave = sum/num; for i=1:num if 0.6<(rad(i)/ave)&&(rad(i)/ave)<1.5 if rad(i)>ave one=one+1; else if 0.93<(rad(i)/ave) && rad(i)<=ave half=half+1; else little = little+1; end end end end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %display results sum=half*0.5+one+little*0.1; one half little sum这段代码什么意思

%清空工作空间中的所有变量和命令窗口内容 clc; clear all; %打开文件选择对话框,选择需要处理的图片 [filename,pathname]=uigetfile({'*.jpg;*.tif;*.png;*gif','all imagine files';'*.*','all files'},'select your photo'); %获取图片路径 path=[pathname,filename]; %读取图片 image=imread(path); %显示图片 imshow(image); %图片处理 %将RGB图像转换为灰度图像 I=rgb2gray(image); %将灰度图像进行滤波操作 I=rangefilt(I); %使用形态学开运算估计背景 background = imopen(I,strel('disk',11)); %从原始图像中减去背景图像 I2 = I-background; %增强对比度 I3 = imadjust(I2); %阈值分割,生成二值图像 bw = imbinarize(I3); %降噪 bw = bwareaopen(bw,160); %进行边缘检测 bw=edge(bw,'canny'); %显示二值图像 imshow(bw); %生成结构元素 se=strel('square',15); %闭运算 bw1=imclose(bw,se); %膨胀 bw2=imdilate(bw1,se); %腐蚀 bw2=imerode(bw2,se); %填充孔洞 bw3=imfill(bw2,'holes'); %显示填充后的二值图像 imshow(bw3); %定义硬币半径取值范围 rmin = 20; rmax = 2500; radiusRange=[rmin rmax]; %使用Hough变换检测圆形目标,返回检测到的圆心坐标和半径大小 [center, rad] = imfindcircles(bw3,radiusRange,'EdgeThreshold',0.13); %显示检测到的圆形目标 imshow(bw3); viscircles(center, rad,'Color','b'); %初始化硬币个数 one=0; half=0; little=0; %对检测到的圆形目标进行分类 [m,n]=size(rad); num=m; i=1; j=num; min=rad(i); max=rad(j); while i<=j if rad(i)<rad(j) if rad(i)<min min=rad(i); else if rad(j)<max max=rad(j); end end else if rad(j)<min min=rad(j); else if rad(i)<max max=rad(i); end end end i=i+1; j=j-1; end sum=0; for i=1:num sum=rad(i)+sum; end ave=(sum-(min+max))/(num-2); for i=1:num if 0.6<(rad(i)/ave)&&(rad(i)/ave)<1.5 if rad(i)>ave one=one+1; else if 0.96<(rad(i)/ave) && rad(i)<=ave half=half+1; else little = little+1; end end end end %计算硬币总价值 sum=half*0.5+one+little*0.1; %显示硬币分类结果 one half little sum 这个程序的不足之处是什么

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