Ergebnisse für
Experimenting with Agentic AI
44%
I am an AI skeptic
0%
AI is banned at work
11%
I am happy with Conversational AI
44%
9 Stimmen
When solving Cody problems, sometimes your solution takes too long — especially if you’re recomputing large arrays or iterative sequences every time your function is called.
The Cody work area resets between separate runs of your code, but within one Cody test suite, your function may be called multiple times in a single session.
This is where persistent variables come in handy.
A persistent variable keeps its value between function calls, but only while MATLAB is still running your function suite.
This means:
- You can cache results to avoid recomputation.
- You can accumulate data across multiple calls.
- But it resets when Cody or MATLAB restarts.
Suppose you’re asked to find the n-th Fibonacci number efficiently — Cody may time out if you use recursion naively. Here’s how to use persistent to store computed values:
function f = fibPersistent(n)
import java.math.BigInteger
persistent F
if isempty(F)
F=[BigInteger('0'),BigInteger('1')];
for k=3:10000
F(k)=F(k-1).add(F(k-2));
end
end
% Extend the stored sequence only if needed
while length(F) <= n
F(end+1)=F(end).add(F(end-1));
end
f = char(F(n+1).toString); % since F(1) is really F(0)
end
%calling function 100 times
K=arrayfun(@(x)fibPersistent(x),randi(10000,1,100),'UniformOutput',false);
K(100)
The fzero function can handle extremely messy equations — even those mixing exponentials, trigonometric, and logarithmic terms — provided the function is continuous near the root and you give a reasonable starting point or interval.
It’s ideal for cases like:
- Solving energy balance equations
- Finding intersection points of nonlinear models
- Determining parameters from experimental data
Example: Solving for Equilibrium Temperature in a Heat Radiation-Conduction Model
Suppose a spacecraft component exchanges heat via conduction and radiation with its environment. At steady state, the power generated internally equals the heat lost:
Given constants:
= 25 W- k = 0.5 W/K
- ϵ = 0.8
- σ = 5.67e−8 W/m²K⁴
- A = 0.1 m²
= 250 K
Find the steady-state temperature, T.
% Given constants
Qgen = 25;
k = 0.5;
eps = 0.8;
sigma = 5.67e-8;
A = 0.1;
Tinf = 250;
% Define the energy balance equation (set equal to zero)
f = @(T) Qgen - (k*(T - Tinf) + eps*sigma*A*(T.^4 - Tinf^4));
% Plot for a sense of where the root lies before implementing
fplot(f, [250 300]); grid on
xlabel('Temperature (K)'); ylabel('f(T)')
title('Energy Balance: Root corresponds to steady-state temperature')
% Use fzero with an interval that brackets the root
T_eq = fzero(f, [250 300]);
fprintf('Steady-state temperature: %.2f K\n', T_eq);
I set my 3D matrix up with the players in the 3rd dimension. I set up the matrix with: 1) player does not hold the card (-1), player holds the card (1), and unknown holding the card (0). I moved through the turns (-1 and 1) that are fixed first. Then cycled through the conditional turns (0) while checking the cards of each player using the hints provided until it was solved. The key for me in solving several of the tests (11, 17, and 19) was looking at the 1's and 0's being held by each player.
sum(cardState==1,3);%any zeros in this 2D matrix indicate possible cards in the solution
sum(cardState==0,3)>0;%the ones in this 2D matrix indicate the only unknown positions
sum(cardState==1,3)|sum(cardState==0,3)>0;%oring the two together could provide valuable information
Some MATLAB Cody problems prohibit loops (for, while) or conditionals (if, switch, while), forcing creative solutions.
One elegant trick is to use nested functions and recursion to achieve the same logic — while staying within the rules.
Example: Recursive Summation Without Loops or Conditionals
Suppose loops and conditionals are banned, but you need to compute the sum of numbers from 1 to n. This is a simple example and obvisously n*(n+1)/2 would be preferred.
function s = sumRecursive(n)
zero=@(x)0;
s = helper(n); % call nested recursive function
function out = helper(k)
L={zero,@helper};
out = k+L{(k>0)+1}(k-1);
end
end
sumRecursive(10)
- The helper function calls itself until the base case is reached.
- Logical indexing into a cell array (k>0) act as an 'if' replacement.
- MATLAB allows nested functions to share variables and functions (zero), so you can keep state across calls.
Tips:
- Replace 'if' with logical indexing into a cell array.
- Replace for/while with recursion.
- Nested functions are local and can access outer variables, avoiding global state.
What a fantastic start to Cody Contest 2025! In just 2 days, over 300 players joined the fun, and we already have our first contest group finishers. A big shoutout to the first finisher from each team:
- Team Creative Coders: @Mehdi Dehghan
- Team Cool Coders: @Pawel
- Team Relentless Coders: @David Hill
- 🏆 First finisher overall: Mehdi Dehghan
Other group finishers: @Bin Jiang (Relentless), @Mazhar (Creative), @Vasilis Bellos (Creative), @Stefan Abendroth (Creative), @Armando Longobardi (Cool), @Cephas (Cool)
Kudos to all group finishers! 🎉
Reminder to finishers: The goal of Cody Contest is learning together. Share hints (not full solutions) to help your teammates complete the problem group. The winning team will be the one with the most group finishers — teamwork matters!
To all players: Don’t be shy about asking for help! When you do, show your work — include your code, error messages, and any details needed for others to reproduce your results.
Keep solving, keep sharing, and most importantly — have fun!
Many MATLAB Cody problems involve recognizing integer sequences.
If a sequence looks familiar but you can’t quite place it, the On-Line Encyclopedia of Integer Sequences (OEIS) can be your best friend.
OEIS will often identify the sequence, provide a formula, recurrence relation, or even direct MATLAB-compatible pseudocode.
Example: Recognizing a Cody Sequence
Suppose you encounter this sequence in a Cody problem:
1, 1, 2, 3, 5, 8, 13, 21, ...
Entering it on OEIS yields A000045 – The Fibonacci Numbers, defined by:
F(n) = F(n-1) + F(n-2), with F(1)=1, F(2)=1
You can then directly implement it in MATLAB:
function F = fibSeq(n)
F = zeros(1,n);
F(1:2) = 1;
for k = 3:n
F(k) = F(k-1) + F(k-2);
end
end
fibSeq(15)
When solving MATLAB Cody problems involving very large integers (e.g., factorials, Fibonacci numbers, or modular arithmetic), you might exceed MATLAB’s built-in numeric limits.
To overcome this, you can use Java’s java.math.BigInteger directly within MATLAB — it’s fast, exact, and often accepted by Cody if you convert the final result to a numeric or string form.
Below is an example of using it to find large factorials.
function s = bigFactorial(n)
import java.math.BigInteger
f = BigInteger('1');
for k = 2:n
f = f.multiply(BigInteger(num2str(k)));
end
s = char(f.toString); % Return as string to avoid overflow
end
bigFactorial(100)
The main round of Cody Contest 2025 kicks off today! Whether you’re a beginner or a seasoned solver, now’s your time to shine.
Here’s how to join the fun:
- Pick your team — choose one that matches your coding personality.
- Solve Cody problems — gain points and climb the leaderboard.
- Finish the Contest Problem Group — help your team win and unlock chances for weekly prizes by finishing the Cody Contest 2025 problem group.
- Share Tips & Tricks — post your insights to win a coveted MathWorks Yeti Bottle.
- Bonus Round — 2 players from each team will be invited to a fun live code-along event!
- Watch Party – join the big watch event to see how top players tackle Cody problems
Contest Timeline:
- Main Round: Nov 10 – Dec 7, 2025
- Bonus Round: Dec 8 – Dec 19, 2025
Big prizes await — MathWorks swag, Amazon gift cards, and shiny virtual badges!
We look forward to seeing you in the contest — learn, compete, and have fun!
Run MATLAB using AI applications by leveraging MCP. This MCP server for MATLAB supports a wide range of coding agents like Claude Code and Visual Studio Code.
Check it out and share your experiences below. Have fun!
GitHub repo: https://github.com/matlab/matlab-mcp-core-server
Yann Debray's blog post: https://blogs.mathworks.com/deep-learning/2025/11/03/releasing-the-matlab-mcp-core-server-on-github/
Hey Relentless Coders! 😎
Let’s get to know each other. Drop a quick intro below and meet your teammates! This is your chance to meet teammates, find coding buddies, and build connections that make the contest more fun and rewarding!
You can share:
- Your name or nickname
- Where you’re from
- Your favorite coding topic or language
- What you’re most excited about in the contest
Let’s make Team Relentless Coders an awesome community—jump in and say hi! 🚀
Welcome to the Cody Contest 2025 and the Relentless Coders team channel! 🎉
You never give up. When a problem gets tough, you dig in deeper. This is your space to connect with like-minded coders, share insights, and help your team win. To make sure everyone has a great experience, please keep these tips in mind:
- Follow the Community Guidelines: Take a moment to review our community standards. Posts that don’t follow these guidelines may be flagged by moderators or community members.
- Ask Questions About Cody Problems: When asking for help, show your work! Include your code, error messages, and any details needed to reproduce your results. This helps others provide useful, targeted answers.
- Share Tips & Tricks: Knowledge sharing is key to success. When posting tips or solutions, explain how and why your approach works so others can learn your problem-solving methods.
- Provide Feedback: We value your feedback! Use this channel to report issues or share creative ideas to make the contest even better.
Have fun and enjoy the challenge! We hope you’ll learn new MATLAB skills, make great connections, and win amazing prizes! 🚀
For the www, uk, and in domains,a generative search answer is available for Help Center searches. Please let us know if you get good or bad results for your searches. Some have pointed out that it is not available in non-english domains. You can switch your country setting to try it out. You can also ask questions in different languages and ask for the response in a different language. I get better results when I ask more specific queries. How is it working for you?
Hello MATLAB Central community,
My name is Yann. And I love MATLAB. I also love Python ... 🐍 (I know, not the place for that).
I recently decided to go down the rabbit hole of AI. So I started benchmarking deep learning frameworks on basic examples. Here is a recording of my experiment:
Happy to engage in the debate. What do you think?
Large Language Models (LLMs) with MATLAB was updated again today to support the newly released OpenAI models GPT-5, GPT-5 mini, GPT-5 nano, GPT-5 chat, o3, and o4-mini. When you create an openAIChat object, set the ModelName name-value argument to "gpt-5", "gpt-5-mini", "gpt-5-nano", "gpt-5-chat-latest", "o4-mini", or "o3".
This is version 4.4.0 of this free MATLAB add-on that lets you interact with LLMs on MATLAB. The release notes are at Release v4.4.0: Support for GPT-5, o3, o4-mini · matlab-deep-learning/llms-with-matlab
Large Languge model with MATLAB, a free add-on that lets you access LLMs from OpenAI, Azure, amd Ollama (to use local models) on MATLAB, has been updated to support OpenAI GPT-4.1, GPT-4.1 mini, and GPT-4.1 nano.
According to OpenAI, "These models outperform GPT‑4o and GPT‑4o mini across the board, with major gains in coding and instruction following. They also have larger context windows—supporting up to 1 million tokens of context—and are able to better use that context with improved long-context comprehension."
What would you build with the latest update?
Provide insightful answers
9%
Provide label-AI answer
9%
Provide answer by both AI and human
21%
Do not use AI for answers
46%
Give a button "chat with copilot"
10%
use AI to draft better qustions
5%
1561 Stimmen
%% 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!
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.
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?
