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Hello Community,
Registration is now open for the MathWorks Automotive Conference 2026 North America. Join industry leaders and MathWorks experts to learn about the latest trends in:
- Software-defined vehicles
- Generative AI
- Virtual vehicles
- AI and machine learning
Event details
- Date: April 28, 2026
- Location: Saint John’s Resort, Plymouth, MI
This conference is a great opportunity to connect with MathWorks and industry peers—and to explore what’s next in automotive engineering. We encourage you to register today.
We hope to see you there.
Hello All,
This is my first post here so I hope its in the right place,
I have built myself a GW consisting of a RAK2245 concentrator and a Raspberry Pi, Also an Arduino end device from this link https://tum-gis-sensor-nodes.readthedocs.io/en/latest/dragino_lora_arduino_shield/README.html
Both projects work fine and connect to TTN whereby packets of data from the end device can be seen in the TTN console.
I now want to create a Webhook in TTN for Thingspeak which would hopefull allow me to see Temperature , Humidity etc in graphical form.
My question, does thingspeak support homebuilt devices or is it focused on comercially built devices ?
I have spent many hours trying to find data hosting site that is comepletely free for a few devices and not to complicated to setup as some seem to be a nightmare. Thanks for any support .
How can I found my license I'd and password, so please provide me my id
A coworker shared with me a hilarious Instagram post today. A brave bro posted a short video showing his MATLAB code… casually throwing 49,000 errors!
Surprisingly, the video went virial and recieved 250,000+ likes and 800+ comments. You really never know what the Instagram algorithm is thinking, but apparently “my code is absolutely cooked” is a universal developer experience 😂
Last note: Can someone please help this Bro fix his code?
In 2025, we saw the growing impact of GenAI on site traffic and user behavior across the entire technical landscape. Amid all this change, MATLAB Central continued to stand out as a trusted home for MATLAB and Simulink users. More than 11 million unique visitors in 2025 came to MATLAB Central to ask questions, share code, learn, and connect with one another.
Let’s celebrate what made 2025 memorable across three key areas: people, content, and events.
People
In 2025, nearly 20,000 contributors participated across the community. We’d like to spotlight a few standout contributors:
- @Sam Chak earned the Most Accepted Answers Badge for both 2024 and 2025. Sam is a rising star in MATLAB Answers with 2,000+ answers and 1,000+ votes.
- @Rodney Tan has been actively contributing files to File Exchange. In 2025, his submissions got almost 20,000 downloads!
- @Dyuman Joshi was recognized as a top contributor on both Cody and Answers. Many may not know that Dyuman is also a Cody moderator, doing tremendous behind-the-scenes moderation work to keep the platform running smoothly.
- A warm welcome to @Steve Eddins, who joined the Community Advisory Board. Steve brings a unique perspective as a former MathWorker and long-time top community contributor.
- Congratulations to @Walter Roberson on reaching 100 followers! MATLAB Central thrives on people-to-people connections, and we’d love to see even more of these relationships grow.
Of course, there are many contributors we didn’t mention here—thank you all for your outstanding contributions and for making the community what it is.
Content
Our high-quality community content not only attracts users but also helps power the broader GenAI ecosystem.
Popular Blog Post & File Exchange Submission
- Zoomed Axes, submitted by @Caleb Thomas, enables zoomed-in views of selected regions in a plot.This submission was featured in the Pick of the Week blog post, “MATLAB Zoomed Axes: Showing zoomed-in regions of a 2D plot,” which generated 5,000 views in just one month.
Popular Discussion Post
- What did MATLAB/Simulink users wait for most in 2025? It's R2025a! “Where is MATLAB R2025a?” became the most-viewed discussion post, with 10,000 views and 30 comments. Thanks for your patience — MATLAB R2025a turned out to be one of the biggest releases we’ve ever delivered.
Most Viewed Question
- “How do I create a for loop in MATLAB?” was the most-viewed community question of the year. It’s a fun reminder that even as MATLAB evolves, the basics remain essential — and always in demand.
Most Voted Poll
- “Did you know there is an official MATLAB certification?”, created by @goc3, was the most-voted poll of 2025.While 50% of respondents voted “No”, it’s exciting to see 3% are certified MATLAB Professionals. Will you be one of them in 2026?
Events
The Cody Contest 2025 brought teams together to tackle challenging but fun Cody problems. During the contest:
- 20,000+ solutions were submitted
- 20+ tips & tricks articles were shared by top players
While the contest has ended, you can still challenge yourself with the fun contest problem group. If you get stuck, the tips & tricks articles are a great resource—and you’ll be amazed by the creativity and skill of the contributors.
Thank you for being part of an incredible 2025. Your curiosity, generosity, and expertise are what make MATLAB Central a trusted home for millions—and we look forward to learning and growing together in 2026.
Is it possible to display a variable value within the ThingSpeak plot area?
I’m currently developing a multi-platform viewer using Flutter to eliminate the hassle of manual channel setup. Instead of adding IDs one by one, the app uses your User API Key to automatically discover and list all your ThingSpeak channels instantly.
Key Highlights (Work in Progress):
- Automatic Sync: All your channels appear in seconds.
- Multi-platform: Built for Web, Android, Windows, and Linux.
- Privacy-Focused: Secure local storage for your API keys.
Give your LLM an easier time looking for information on mathworks.com: point it to the recently released llms.txt files. The top-level one is www.mathworks.com/llms.txt, release changes use www.mathworks.com/help/relnotes. How does it work for you??
I can't believe someone put time into this ;-)
Hi everyone
I've been using ThingSpeak for several years now without an issue until last Thursday.
I have four ThingSpeak channels which are used by three Arduino devices (in two locations/on two distinct networks) all running the same code.
All three devices stopped being able to write data to my ThingSpeak channels around 17:00 CET on 4 Dec and are still unable to.
Nothing changed on this side, let alone something that would explain the problem.
I would note that data can still be written to all the channels via a browser so there is no fundamental problem with the channels (such as being full).
Since the above date and time, any HTTP/1.1 'update' (write) requests via the REST API (using both simple one-write GET requests or bulk JSON POST requests) are timing out after 5 seconds and no data is being written. The 5 second timeout is my Arduino code's default, but even increasing it to 30 seconds makes no difference. Before all this, responses from ThingSpeak were sub-second.
I have recompiled the Arduino code using the latest libraries and that didn't help.
I have tested the same code again another random api (api.ipify.org) and that works just fine.
Curl works just fine too, also usng HTTP/1.1
So the issue appears to be something particular to the combination of my Arduino code *and* the ThingSpeak environment, where something changed on the ThingSpeak end at the above date and time.
If anyone in the community has any suggestions as to what might be going on, I would greatly appreciate the help.
Peter
The formula comes from @yuruyurau. (https://x.com/yuruyurau)
digital life 1
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:2e4;
x = mod(i, 100);
y = floor(i./100);
k = x./4 - 12.5;
e = y./9 + 5;
o = vecnorm([k; e])./9;
while true
t = t + pi/90;
q = x + 99 + tan(1./k) + o.*k.*(cos(e.*9)./4 + cos(y./2)).*sin(o.*4 - t);
c = o.*e./30 - t./8;
SHdl.XData = (q.*0.7.*sin(c)) + 9.*cos(y./19 + t) + 200;
SHdl.YData = 200 + (q./2.*cos(c));
drawnow
end
digital life 2
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:1e4;
x = i;
y = i./235;
e = y./8 - 13;
while true
t = t + pi/240;
k = (4 + sin(y.*2 - t).*3).*cos(x./29);
d = vecnorm([k; e]);
q = 3.*sin(k.*2) + 0.3./k + sin(y./25).*k.*(9 + 4.*sin(e.*9 - d.*3 + t.*2));
SHdl.XData = q + 30.*cos(d - t) + 200;
SHdl.YData = 620 - q.*sin(d - t) - d.*39;
drawnow
end
digital life 3
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 1, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:1e4;
x = mod(i, 200);
y = i./43;
k = 5.*cos(x./14).*cos(y./30);
e = y./8 - 13;
d = (k.^2 + e.^2)./59 + 4;
a = atan2(k, e);
while true
t = t + pi/20;
q = 60 - 3.*sin(a.*e) + k.*(3 + 4./d.*sin(d.^2 - t.*2));
c = d./2 + e./99 - t./18;
SHdl.XData = q.*sin(c) + 200;
SHdl.YData = (q + d.*9).*cos(c) + 200;
drawnow; pause(1e-2)
end
digital life 4
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 1, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:4e4;
x = mod(i, 200);
y = i./200;
k = x./8 - 12.5;
e = y./8 - 12.5;
o = (k.^2 + e.^2)./169;
d = .5 + 5.*cos(o);
while true
t = t + pi/120;
SHdl.XData = x + d.*k.*sin(d.*2 + o + t) + e.*cos(e + t) + 100;
SHdl.YData = y./4 - o.*135 + d.*6.*cos(d.*3 + o.*9 + t) + 275;
SHdl.CData = ((d.*sin(k).*sin(t.*4 + e)).^2).'.*[1,1,1];
drawnow;
end
digital life 5
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 1, 'filled','o','w',...
'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 0:1e4;
x = mod(i, 200);
y = i./55;
k = 9.*cos(x./8);
e = y./8 - 12.5;
while true
t = t + pi/120;
d = (k.^2 + e.^2)./99 + sin(t)./6 + .5;
q = 99 - e.*sin(atan2(k, e).*7)./d + k.*(3 + cos(d.^2 - t).*2);
c = d./2 + e./69 - t./16;
SHdl.XData = q.*sin(c) + 200;
SHdl.YData = (q + 19.*d).*cos(c) + 200;
drawnow;
end
digital life 6
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 1:1e4;
y = i./790;
k = y; idx = y < 5;
k(idx) = 6 + sin(bitxor(floor(y(idx)), 1)).*6;
k(~idx) = 4 + cos(y(~idx));
while true
t = t + pi/90;
d = sqrt((k.*cos(i + t./4)).^2 + (y/3-13).^2);
q = y.*k.*cos(i + t./4)./5.*(2 + sin(d.*2 + y - t.*4));
c = d./3 - t./2 + mod(i, 2);
SHdl.XData = q + 90.*cos(c) + 200;
SHdl.YData = 400 - (q.*sin(c) + d.*29 - 170);
drawnow; pause(1e-2)
end
digital life 7
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.4);
t = 0;
i = 1:1e4;
y = i./345;
x = y; idx = y < 11;
x(idx) = 6 + sin(bitxor(floor(x(idx)), 8))*6;
x(~idx) = x(~idx)./5 + cos(x(~idx)./2);
e = y./7 - 13;
while true
t = t + pi/120;
k = x.*cos(i - t./4);
d = sqrt(k.^2 + e.^2) + sin(e./4 + t)./2;
q = y.*k./d.*(3 + sin(d.*2 + y./2 - t.*4));
c = d./2 + 1 - t./2;
SHdl.XData = q + 60.*cos(c) + 200;
SHdl.YData = 400 - (q.*sin(c) + d.*29 - 170);
drawnow; pause(5e-3)
end
digital life 8
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl{6} = [];
for j = 1:6
SHdl{j} = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.3);
end
t = 0;
i = 1:2e4;
k = mod(i, 25) - 12;
e = i./800; m = 200;
theta = pi/3;
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
while true
t = t + pi/240;
d = 7.*cos(sqrt(k.^2 + e.^2)./3 + t./2);
XY = [k.*4 + d.*k.*sin(d + e./9 + t);
e.*2 - d.*9 - d.*9.*cos(d + t)];
for j = 1:6
XY = R*XY;
SHdl{j}.XData = XY(1,:) + m;
SHdl{j}.YData = XY(2,:) + m;
end
drawnow;
end
digital life 9
clc; clear
figure('Position',[300,50,900,900], 'Color','k');
axes(gcf, 'NextPlot','add', 'Position',[0,0,1,1], 'Color','k');
axis([0, 400, 0, 400])
SHdl{14} = [];
for j = 1:14
SHdl{j} = scatter([], [], 2, 'filled','o','w', 'MarkerEdgeColor','none', 'MarkerFaceAlpha',.1);
end
t = 0;
i = 1:2e4;
k = mod(i, 50) - 25;
e = i./1100; m = 200;
theta = pi/7;
R = [cos(theta) -sin(theta); sin(theta) cos(theta)];
while true
t = t + pi/240;
d = 5.*cos(sqrt(k.^2 + e.^2) - t + mod(i, 2));
XY = [k + k.*d./6.*sin(d + e./3 + t);
90 + e.*d - e./d.*2.*cos(d + t)];
for j = 1:14
XY = R*XY;
SHdl{j}.XData = XY(1,:) + m;
SHdl{j}.YData = XY(2,:) + m;
end
drawnow;
end
If you haven't solved the problem yet, below hints guide how the algorithm should be implemented and clarify subtle rules that are easy to miss.
1. Shield is ONLY defended in HOME matches of the CURRENT holder - Even if a team beats the Shield holder in an away match, that does NOT count as a Shield defense.
2. A team defends the Shield ONLY when:
> They currently hold it.
> They are home team in that match
3. Shield transfer happens ONLY if the HOLDER plays a home match AND loses - A team may lose an away match — no effect.
4. The output ALWAYS includes the initial holder as the first row.
5. Defenses count resets for each new holder. - Every holder accumulates their own count until they lose it at home.
6. Match numbers are 1-indexed in the input, but “0” is used for initial state - The first real match is Match 1, but the output starts with Match 0.
7. Output row is created ONLY WHEN SHIELD CHANGES HANDS - This is an important hidden detail. A new row is appended, When the current holder loses a home match → Shield taken by visitor. If no loss at home occurs after that → no new row until next change.
8. The last holder’s defense count goes until the season ends - Even if they lose away later.
9. If a holder never gets a home match, defenses = 0.
10. In case the holder loses their very first home match → defenses = 0.
11. Shield changes only on HOME LOSS, not on a draw.
I hope above hints will help you in solving the problem.
Thanks and Regards,
Dev
Hello,
I have Arduino DIY Geiger Counter, that uploads data to my channel here in ThingSpeak (3171809), using ESP8266 WiFi board. It sends CPM values (counts per minute), Dose, VCC and Max CPM for 24h. They are assignet to Field from 1 to 4 respectively. How can I duplicate Field 1, so I could create different time chart for the same measured unit? Or should I duplicate Field 1 chart, and how? I tried to find the answer here in the blog, but I couldn't.
I have to say that I'm not an engineer or coder, just can simply load some Arduino sketches and few more things, so I'll be very thankfull if someone could explain like for non-IT users.
Regards,
Emo
Many MATLAB Cody problems involve solving congruences, modular inverses, Diophantine equations, or simplifying ratios under constraints. A powerful tool for these tasks is the Extended Euclidean Algorithm (EEA), which not only computes the greatest common divisor, gcd(a,b), but also provides integers x and y such that: a*x + b*y = gcd(a,b) - which is Bezout's identity.
Use of the Extended Euclidean Algorithm is very using in solving many different types of MATLAB Cody problems such as:
- Computing modular inverses safely, even for very large numbers
- Solving linear Diophantine equations
- Simplifing fractions or finding nteger coefficients without using symbolic tools
- Avoiding loops (EEA can be implemented recursively)
Below is a recursive implementation of the EEA.
function [g,x,y] = egcd(a,b)
% a*x + b*y = g [gcd(a,b)]
if b == 0
g = a; x = 1; y = 0;
else
[g, x1, y1] = egcd(b, mod(a,b));
x = y1;
y = x1 - floor(a/b)*y1;
end
end
Problem:
Given integers a and m, return the modular inverse of a (mod m).
If the inverse does not exist, return -1.
function inv = modInverse(a,m)
[g,x,~] = egcd(a,m);
if g ~= 1 % inverse doesn't exist
inv = -1;
else
inv = mod(x,m); % Bézout coefficient gives the inverse
end
end
%find the modular inverse of 19 (mod 5)
inv=modInverse(19,5)
Congratulations to all the Relentless Coders who have completed the problem set. I hope you weren't too busy relentlessly solving problems to enjoy the silliness I put into them.
If you've solved the whole problem set, don't forget to help out your teammates with suggestions, tips, tricks, etc. But also, just for fun, I'm curious to see which of my many in-jokes and nerdy references you noticed. Many of the problems were inspired by things in the real world, then ported over into the chaotic fantasy world of Nedland.
I guess I'll start with the obvious real-world reference: @Ned Gulley (I make no comment about his role as insane despot in any universe, real or otherwise.)
Hi Everyone!
As this is the most difficult question in problem group "Cody Contest 2025". To solve this problem, It is very important to understand all the hidden clues in the problem statement. Because everything is not directly visible.
For those who tried the problem, but were not able to solve. You might have missed any of the below hints -
- “The other players do not get to see which card has been shown, but they do know which three cards were asked for and that the player asked had one of them.” - Even when the card identity isn’t revealed (result = 0), you still gain partial knowledge — the asked player must have at least one of those three cards, meaning you can mark other players as not having all three simultaneously.
- "If it is your turn, you know the exact identity of that card" - You only know the exact shown card when result = 1, 2, or 3 — and it must be your turn. If someone else asked (even if you know result = 0), you don’t know which one was shown. So the meaning of result depends on whose turn it was, which is implicit — MATLAB code must assume that turns alternate 1→m→1, so your turn index is determined by (t-1) mod m + 1 == pnum.
- "Any leftover cards are placed face-up so that all players can see them" - These cards (commoncards) are not in anyone’s hand and cannot be in the envelope. So they’re not just visible — they’re logical constraints to eliminate from deduction.
- “It may be possible to determine the solution from less information than is given, but the information given will always be sufficient.”
- "Turn order is implied, not given explicitly" - Players take turns in order (1 to m, and back to 1).
On considering all the clues and constraints in the question, you will definitely be able to card for each category present in envelope.
I hope above clues will be useful for you.
Thank you, wishing you the success!
Regards,
Dev
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);
Pure Matlab
82%
Simulink
18%
11 Stimmen
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
