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I have written, tested, and prepared a function with four subsunctions on my computer for solving one of the problems in the list of Cody problems in MathWorks in three days. Today, when I wanted to upload or copy paste the codes of the function and its subfunctions to the specified place of the problem of Cody page, I do not see a place to upload it, and the ability to copy past the codes. The total of the entire codes and their documentations is about 600 lines, which means that I cannot and it is not worth it to retype all of them in the relevent Cody environment after spending a few days. I would appreciate your guidance on how to enter the prepared codes to the desired environment in Cody.
xingxingcui
xingxingcui
Última actividad el 29 de Mzo. de 2025

看到知乎有用Origin软件绘制3D瀑布图,觉得挺美观的,突然也想用MATLAB复现一样的图,借助ChatGPT,很容易写出代码,相对Origin软件,无需手动干预调整图像属性,代码控制性强:
%% 清理环境
close all; clear; clc;
%% 模拟时间序列
t = linspace(0,12,200); % 时间从 0 到 12,分 200 个点
% 下面构造一些模拟的"峰状"数据,用于演示
% 你可以根据需要替换成自己的真实数据
rng(0); % 固定随机种子,方便复现
baseIntensity = -20; % 强度基线(z 轴的最低值)
numSamples = 5; % 样本数量
yOffsets = linspace(20,140,numSamples); % 不同样本在 y 轴上的偏移
colors = [ ...
0.8 0.2 0.2; % 红
0.2 0.8 0.2; % 绿
0.2 0.2 0.8; % 蓝
0.9 0.7 0.2; % 金黄
0.6 0.4 0.7]; % 紫
% 构造一些带多个峰的模拟数据
dataMatrix = zeros(numSamples, length(t));
for i = 1:numSamples
% 随机峰参数
peakPositions = randperm(length(t),3); % 三个峰位置
intensities = zeros(size(t));
for pk = 1:3
center = peakPositions(pk);
width = 10 + 10*rand; % 峰宽
height = 100 + 50*rand; % 峰高
% 高斯峰
intensities = intensities + height*exp(-((1:length(t))-center).^2/(2*width^2));
end
% 再加一些小随机扰动
intensities = intensities + 10*randn(size(t));
dataMatrix(i,:) = intensities;
end
%% 开始绘图
figure('Color','w','Position',[100 100 800 600],'Theme','light');
hold on; box on; grid on;
for i = 1:numSamples
% 构造 fill3 的多边形顶点
xPatch = [t, fliplr(t)];
yPatch = [yOffsets(i)*ones(size(t)), fliplr(yOffsets(i)*ones(size(t)))];
zPatch = [dataMatrix(i,:), baseIntensity*ones(size(t))];
% 使用 fill3 填充面积
hFill = fill3(xPatch, yPatch, zPatch, colors(i,:));
set(hFill,'FaceAlpha',0.8,'EdgeColor','none'); % 调整透明度、去除边框
% 在每条曲线尾部标注 Sample i
text(t(end)+0.3, yOffsets(i), dataMatrix(i,end), ...
['Sample ' num2str(i)], 'FontSize',10, ...
'HorizontalAlignment','left','VerticalAlignment','middle');
end
%% 坐标轴与视角设置
xlim([0 12]);
ylim([0 160]);
zlim([-20 350]);
xlabel('Time (sec)','FontWeight','bold');
ylabel('Frequency (Hz)','FontWeight','bold');
zlabel('Intensity','FontWeight','bold');
% 设置刻度(根据需要微调)
set(gca,'XTick',0:2:12, ...
'YTick',0:40:160, ...
'ZTick',-20:40:200);
% 设置视角(az = 水平旋转,el = 垂直旋转)
view([211 21]);
% 让三维坐标轴在后方
set(gca,'Projection','perspective');
% 如果想去掉默认的坐标轴线,也可以尝试
% set(gca,'BoxStyle','full','LineWidth',1.2);
%% 可选:在后方添加一个浅色网格平面 (示例)
% 这个与题图右上方的网格类似
[Xplane,Yplane] = meshgrid([0 12],[0 160]);
Zplane = baseIntensity*ones(size(Xplane)); % 在 Z = -20 处画一个竖直面的框
surf(Xplane, Yplane, Zplane, ...
'FaceColor',[0.95 0.95 0.9], ...
'EdgeColor','k','FaceAlpha',0.3);
%% 进一步美化(可根据需求调整)
title('3D Stacked Plot Example','FontSize',12);
constantplane("x",12,FaceColor=rand(1,3),FaceAlpha=0.5);
constantplane("y",0,FaceColor=rand(1,3),FaceAlpha=0.5);
constantplane("z",-19,FaceColor=rand(1,3),FaceAlpha=0.5);
hold off;
Have fun! Enjoy yourself!
David
David
Última actividad el 24 de Mzo. de 2025

We are excited to announce the first edition of the MathWorks AI Challenge. You’re invited to submit innovative solutions to challenges in the field of artificial intelligence. Choose a project from our curated list and submit your solution for a chance to win up to $1,000 (USD). Showcase your creativity and contribute to the advancement of AI technology.
I've been trying this problem a lot of time and i don't understand why my solution doesnt't work.
In 4 tests i get the error Assertion failed but when i run the code myself i get the diag and antidiag correctly.
function [diag_elements, antidg_elements] = your_fcn_name(x)
[m, n] = size(x);
% Inicializar los vectores de la diagonal y la anti-diagonal
diag_elements = zeros(1, min(m, n));
antidg_elements = zeros(1, min(m, n));
% Extraer los elementos de la diagonal
for i = 1:min(m, n)
diag_elements(i) = x(i, i);
end
% Extraer los elementos de la anti-diagonal
for i = 1:min(m, n)
antidg_elements(i) = x(m-i+1, i);
end
end
imad
imad
Última actividad el 14 de Feb. de 2025

Simulink has been an essential tool for modeling and simulating dynamic systems in MATLAB. With the continuous advancements in AI, automation, and real-time simulation, I’m curious about what the future holds for Simulink.
What improvements or new features do you think Simulink will have in the coming years? Will AI-driven modeling, cloud-based simulation, or improved hardware integration shape the next generation of Simulink?
You've probably heard about the DeepSeek AI models by now. Did you know you can run them on your own machine (assuming its powerful enough) and interact with them on MATLAB?
In my latest blog post, I install and run one of the smaller models and start playing with it using MATLAB.
Larger models wouldn't be any different to use assuming you have a big enough machine...and for the largest models you'll need a HUGE machine!
Even tiny models, like the 1.5 billion parameter one I demonstrate in the blog post, can be used to demonstrate and teach things about LLM-based technologies.
Have a play. Let me know what you think.
Brent
Brent
Última actividad el 18 de Feb. de 2025

My following code works running Matlab 2024b for all test cases. However, 3 of 7 tests fail (#1, #4, & #5) the QWERTY Shift Encoder problem. Any ideas what I am missing?
Thanks in advance.
keyboardMap1 = {'qwertyuiop[;'; 'asdfghjkl;'; 'zxcvbnm,'};
keyboardMap2 = {'QWERTYUIOP{'; 'ASDFGHJKL:'; 'ZXCVBNM<'};
if length(s) == 0
se = s;
end
for i = 1:length(s)
if double(s(i)) >= 65 && s(i) <= 90
row = 1;
col = 1;
while ~strcmp(s(i), keyboardMap2{row}(col))
if col < length(keyboardMap2{row})
col = col + 1;
else
row = row + 1;
col = 1;
end
end
se(i) = keyboardMap2{row}(col + 1);
elseif double(s(i)) >= 97 && s(i) <= 122
row = 1;
col = 1;
while ~strcmp(s(i), keyboardMap1{row}(col))
if col < length(keyboardMap1{row})
col = col + 1;
else
row = row + 1;
col = 1;
end
end
se(i) = keyboardMap1{row}(col + 1);
else
se(i) = s(i);
end
% if ~(s(i) = 65 && s(i) <= 90) && ~(s(i) >= 97 && s(i) <= 122)
% se(i) = s(i);
% end
end
Is it possible to differenciate the input, output and in-between wires by colors?
Bob
Bob
Última actividad el 4 de Dic. de 2024

I was curious to startup your new AI Chat playground.
The first screen that popped up made the statement:
"Please keep in mind that AI sometimes writes code and text that seems accurate, but isnt"
Can someone elaborate on what exactly this means with respect to your AI Chat playground integration with the Matlab tools?
Are there any accuracy metrics for this integration?
Local large language models (LLMs), such as llama, phi3, and mistral, are now available in the Large Language Models (LLMs) with MATLAB repository through Ollama™!
Read about it here:
goc3
goc3
Última actividad el 3 de Dic. de 2024

I was browsing the MathWorks website and decided to check the Cody leaderboard. To my surprise, William has now solved 5,000 problems. At the moment, there are 5,227 problems on Cody, so William has solved over 95%. The next competitor is over 500 problems behind. His score is also clearly the highest, approaching 60,000.
Please take a moment to congratulate @William.
J.K043006
J.K043006
Última actividad el 29 de Ag. de 2024

I've been working on some matrix problems recently(Problem 55225)
and this is my code
It turns out that "Undefined function 'corr' for input arguments of type 'double'." However, should't the input argument of "corr" be column vectors with single/double values? What's even going on there?
Daniel Bending
Daniel Bending
Última actividad el 20 de Ag. de 2024

Hi All,
I'm currently verifying a global sensitivity analysis done in SimBiology and I'm a touch confused. This analysis was run with every parameter and compartment volume in the model. To my understanding the fraction of unexplained variance is 1 - the sum of the first order variances, therefore if the model dynamics are dominated by interparameter effects you might see a higher fraction of unexplained variance. In this analysis however, as the attached figure shows (with input at t=20 minutes), the most sensitive four parameters seem to sum, in first order sensitivities to roughly one at each time point and the total order sensitivies appear nearly identical. So how is the fraction of unexplained variance near one?
Thank you for your help!
supercomputers
supercomputers
Última actividad el 6 de Nov. de 2024

hello i found the following tools helpful to write matlab programs. copilot.microsoft.com chatgpt.com/gpts gemini.google.com and ai.meta.com. thanks a lot and best wishes.
Check out the LLMs with MATLAB project on File Exchange to access Large Language Models from MATLAB.
Along with the latest support for GPT-4o mini, you can use LLMs with MATLAB to generate images, categorize data, and provide semantic analyis.
Run it now by clicking Open in MATLAB Online, signing in, and using your API Key from OpenAI.
function ans = your_fcn_name(n)
n;
j=sum(1:n);
a=zeros(1,j);
for i=1:n
a(1,((sum(1:(i-1))+1)):(sum(1:(i-1))+i))=i.*ones(1,i);
end
disp
Marisa
Marisa
Última actividad el 26 de Ag. de 2024

I am trying to earn my Intro to MATLAB badge in Cody, but I cannot click the Roll the Dice! problem. It simply is not letting me click it, therefore I cannot earn my badge. Does anyone know who I should contact or what to do?
Kalhara
Kalhara
Última actividad el 19 de Jun. de 2024

What do you think about the NVIDIA's achivement of becoming the top giant of manufacturing chips, especially for AI world?
Hi to everyone!
To simplify the explanation and the problem, I simulated the kinetics of an irreversible first-order reaction, A -> B. I implemented it in two independent compartments, R and P. I simulated the effect of a dilution in R by doubling at t= 0,1 the R volume. I programmed in P that, at t = 0.1, the instantaneous concentration of A and B would be reduced by half. I am sending an attach with the implementation of these simulations in the Simbiology interface.
When the simulations of the two compartments are plotted, it can be seen that the responses are not equal. That is, from t = 0.1 s, the reaction follow an exponential function in R with half of the initial amplitude and half of the initial value of k1. That is, the relaxation time is doubled. Meanwhile, in P, from t = 0.1, the reaction follows exponential kinetics with half the amplitude value but maintaining the initial value of k = 10. Without a doubt, the correct simulation is the latter (compartment P) where only the effect is observed in the amplitude and not in the relaxation time. Could you tell me what the error is that makes these kinetics that should be equal not be?
Thank you in advance!
Luis B.